Elementa Geometriae

[English] Latin	English [Latin][¶]
The most renowned work of The Elements by Euclid of Megara [sic] together with the commentary of the most perspicacious Campanus on the art of geometry happily begins:	Praeclarissimum opus elementorum Euclidis Megarensis [sic] una cum commentis Campani perspicacissimi in artem geometriam incipit feliciter:
A point is that of which there is no part.	Punctus est cuius pars non est.
¶ A line is length without breadth, of whose extremities are in fact two points.	¶ Linea est longitudo sine latitudine cuius quidem extremitates sunt duo puncta.
¶ A right line is the shortest extension from one point to another, each of them admitting at their extremities.	¶ Linea recta est ab uno puncto ad alium brevissima extensio in extremitates suas utrumque eorum recipiens.
¶ A superficies is that which merely has length and breadth, and whose termini are indeed lines.	¶ Superficies est qui longitudinem et latitudinem tantum habet: cuius termini quidem sunt lineae.
¶ A plane superficies is the extension from one line to another, each admitting at their extremities.	¶ Superficies plana est ab una linea ad aliam extensio in extremitates suas recipiens.
¶ A plane angle is the contact of two distinct lines, the expansion of which is on a superficies and application thereof is not direct.	¶ Angulus planus est duarum linearum alterius contactus: quarum expansio est super superficiem applicationque non directa.
¶ However, when two right lines contain an angle it is called a right-lined angle.	¶ Quando autem angulum continent duae lineae recte rectilineus angulus nominatur.
¶ When a right line stands over a right line and the two angles are both equal, each of them will be right.	¶ Quando recta linea super rectam steterit duoque anguli utrobique fuerint aequales eorum uterque rectus erit.
¶ And the line standing over the line is called perpendicular to that which it is standing.	¶ Lineaque lineae superstans ei cui superstat perpendicularis vocatur.
¶ In particular, it is said that an angle that is greater than right is obtuse.	¶ Angulus vero qui recto maior est obtusus dicitur.
¶ But an angle smaller than right is called acute.	¶ Angulus vero minor recto acutus appellatur.
¶ A terminus is that which is the end of every thing.	¶ Terminus est quod uniuscuiusque finis est.
¶ A figure is that contained by a terminus and/or termini.	¶ Figura est quae termino vel terminis continentur.
¶ A circle is a plane figure enclosed by just one line, which is called the circumference, in whose middle is a point from which all right lines exiting to the circumference are equal to one another. And this point in particular is called the center of the circle.	¶ Circulus est figura plana una quidem linea contenta: quae circumferentia nominatur: in cuius medio punctus est a quo omnes lineae rectae ad circumferentiam exeuntes sibi invicem sunt aequales. Et hic quidem punctus centrum circuli dicitur.
¶ The diameter of a circle is a right line that divides the circle into two halves by passing over its center, applying to the extremities of its circumference.	¶ Diameter circuli est linea recta quae super eius centrum transiens extremitatesque suas circumferentiae applicans circulum in duo media dividit.
¶ A semicircle is a plane figure enclosed within the diameter of a circle and half the circumference.	¶ Semicirculus est figura plana diametro circuli et medietate circumferentiae contenta.
¶ A portion of a circle is a plane figure enclosed within a right line and a part of the circumference, and is in fact either larger or smaller than a semicircle.	¶ Portio circuli est figura plana recta linea et parte circumferentiae contenta: semicirculo quidem aut maior aut minor.
¶ Right-lined figures are those contained by right lines: some of which are trilaterals, which are within three right lines; some quadrilaterals, which are within four right lines; some multilateral, which are those contained within more than four right lines.	¶ Rectilineae figurae sunt quae rectis lineis continentur quarum quaedam trilaterae quae tribus rectis lineis: quaedam quadrilaterae quae quatuor rectis lineis: quaedam multilaterae quae pluribus quae quatuor rectis lineis continentur.
¶ Of the trilateral figures: one is the triangle, having three equal sides; another is the triangle having two equal sides; and another is the triangle of three inequal sides. Of these, again, one is the orthogon, having, namely, one right angle; another is the amblygon, having any obtuse angle; and another is the oxygon, in which there are three acute angles.	¶ Figurarum trilaterarum: alia est triangulus habens tria latera aequalia. Alia triangulus duo habens aequalia latera. Alia triangulus trium inaequalium laterum. Harum iterum alia est orthogonium: unum scilicet rectum angulum habens. Alia est ambligonium aliquem obtusum angulum habens. Alia est oxigonium: in quae tres anguli sunt acuti.
¶ Moreover of the quadrilateral figures: one is the quadrate, which is of equal sides and yet rectangular; another is the long tetragon, which is a rectangular figure but is not equilateral; another is the helmuaym, which is equilateral but is not rectangular.	¶ Figurarum autem quadrilaterarum. Alia est quadratum quod est aequilaterum atque rectangulum. Alia est tetragonus longus: quae est figura rectangula: sed aequilateral non est. Alia est helmuaym: quae est aequilatera: sed rectangula non est.
¶ And another is similar to the helmuaym, which has equal opposite sides and equal opposite angles (yet the aforesaid is neither contained within right angles nor within equal sides). However beyond these, all quadrilateral figures are called helmuariphae.	¶ Alia est similis helmuaym quae opposita latera habet aequalia atque oppositos angulos aequales: idem temen nec rectis angulis nec aequis lateribus continetur. Praeter has autem omnes quadrilaterae figurae helmuariphae nominantur.
¶ Equidistant lines are those which are assembled upon the same surface and one or the other extended in either direction shall not come together, even if extended neverendingly.	¶ Aequidistantes lineae sunt quae in eadem superficie collocatae atque in alterutram partem protractae non conveniunt etiam si in infinium protrahantur.
There are five petitions:	Petitiones sunt quinque.
¶ From whatever point to whatever point, to draw a right line and to then extend that definite line right and continuous manner howsoever is pleasing.	¶ A quolibet puncto in quemlibet punctum rectam lineam ducere atque lineam definitam in continuum rectumque quantumlibet protrahere.
¶ Upon any center, draw a circle occupying a space however great as is pleasing.	¶ Super centrum quodlibet quantumlibet occupando spacium: circulum designare:
¶ All right angles are to be equal to one another.	¶ Omnes rectos angulos sibi invicem esse aequales.
¶ If a right line falls over two right lines and the two angles from one side are smaller than two right angles, those two lines extending in the same direction will undoubtedly proceed to join together.	¶ Si linea recta super duas lineas rectas ceciderit duoque anguli ex una parte duobus rectis angulis minores fuerint istas duas lineas in eandem partem protractas proculdubio coniunctim ire.
¶ Two right lines are to define no superficies.	¶ Duas lineas rectas superficiem nullam concludere.
These are the common conceptions of the mind:	Communes animi conceptiones sunt haec.
¶ Those which are equal to one and the same are mutually equal to each other.	¶ Quae uni et eidem sunt aequalia et sibi invicem sunt aequalia.
¶ And if to equals, equals are added, the totals will also become equal.	¶ Et si aequalibus aequalia addantur tota quoque fient aequalia.
¶ And if from equals, equals are ablated, those left remaining will be equal.	¶ Et si ab aequalibus aequalia auferantur quae relinquuntur erunt aequalia.
¶ And if from unequals equals are removed, those left remaining will be inequal.	¶ Et si ab inaequalibus aequalia demas quae relinquuntur erunt inaequalia.
¶ And if to unequals you add equals, those likewise will made inequal.	¶ Et si inaequalibus aequalia addas ipsa quoque fient inaequalia.
¶ If two things are equal to one, they will mutually equal one another.	¶ Si fuerint duae res uni aequales ipsae sibi invicem erunt aequales.
¶ If there are two things, of which each of one is half the same, one will equal the other.	¶ Si fuerint duae res quarum utraque unius eiusdem fuerit dimidium utraque erit aequalis alteri.
¶ If anything is placed over another and applied to it, and neither one shall the other exceed, they will mutually equal one another..	¶ Si aliqua res alicui superponatur appliceturque ei nec excedat altera alteram: illae sibi invicem erunt aequales.
¶ Every whole is greater than its part.	¶ Omne totum est maius sua parte.
It is understood, however, that beyond these conceptions of the mind, Euclid alternatively omitted many other common knowledges, which are incomprehensible in number; of which this is one:	Sciendum est autem quod praeter has animi conceptiones: sive communes scientias multas alias quae numero sunt incomprehensibiles praetermisit Euclides: quarum haec est una.
¶ If two equal quantities are compared to any third of the same kind, from that third both will together be either equally greater than, equally less than, or simultaneously equal.	¶ Si duae quantitates aequales ad quamlibet tertiam eiusdem generis comparentutur simul erunt ambae illa tertia aut aequae maiores: aut aeque minores: aut simul aequales.
¶ And another thing:	¶ Item alia
How much any quantity is to any other of the same kind is to be of such measure as to any third to any fourth of the same kind, in continuous quantities. This is universally true, whether the preceding will be greater than the succeeding or less than, because magnitude decreases neverendingly. Though in numbers, it is not so; but if the first will be a submultiple of the second, any third will be equally a submultiple of any fourth, since a number increases neverendingly just as magnitude decreases neverendingly.	Quanta est aliqua quantitas ad quamlibet aliam eiusdem generis tantam esse quamlibet tertiam ad aliquam quartani eiusdem generis in quantitatibus continuis: hoc universaliter verum est sive antecedentes maiores fuerint consequentibus sive minores. magnitudo enim decrescit in infinitum. in numeris autem non sic: sed si fuerit primus submultiplex secundi: erit quilibet tertius aeque submultiplex alicuius quarti: quoniam numerus crescit in infinitum: sicut magnitudo in infinitum minuitur.
Proposition 1	Propositio .1.
[English] Latin	English [Latin][¶]
Assemble an equilateral triangle over a given right line.	Triangulum aequilaterum supra datam lineam rectam collocare.
¶ It will be over the given right line a.b that I wish to construct an equilateral triangle on one extremity of it, namely, at point a. There I will place the immobile foot of the compass and I will extend the other movable foot up 'til b, and I will draw out the circle c.b.d.f following the length of that given line by the second petition. In turn, its other end, namely point b, I will make the center. And by the same petition and according to same the extent, I will draw the circle c.a.d.h, wherewith the circles will intersect themselves at two points, which are c and d, and one of the two sections, as section d, I will connect to both ends of the given line with the extended lines d.a and d.b, by means of the first petition. Therefore, because the lines a.d and a.b are extended from point a, which is the center of the circle c.b.d, up to its circumference, they will be equal through the definition of a circle.	¶ Esto data linea recta .a.b. volo super ipsam triangulum aequilaterum constituere super alteram eius extremitatem scilicet in puncto .a. ponam pedem circin immobilem: et alterum pedem mobilem extendam usque ad .b. et describam secundum quantitatem ipsius lineae datae per secundam petitionem circulum .c.b.d.f. rursus alteram eius extremitatem, scilicet, punctum .b. faciam centrum: et per eandem petitionem et secundum eiusdem quantitatem lineabo circulum .c.a.d.h. qui circuli intersecabunt se in duobus punctis quae sint .c.d. et alteram duarum sectionum: sicut sectionem .d. continuabo cum ambabus extremitatibus datae lineae protractis lineis .d.a.d.b. per primam petitionem: quia ergo a puncto .a. quod est centrum circuli .c.b.d. protractae sunt lineae .a.d. et .a.b. usque ad eius circumferentiam ipsae erunt aequales per diffinitionem circuli.
And likewise, because the lines b.a and a.d are extended from point b, which is the center of the circle c.a.d, up to its circumference, they will also be equal, therefore, because each of the two lines a.d and b.d equal the line a.b as has been demonstrated, they will be equal among themselves through the first conception, thus, we have assembled an equilateral triangle over the given line, which is the intent.	Similiter quoque: quia a puncto .b. quod est centrum circuli .c.a.d. protractae sunt lineae .b.a. et .a.d. usque ad eius circumferentiam ipsae erunt etiam aequales, quia ergo utraque duarum linearum .a.d.b.d. aequalis est linea .a.b. ut probatum est: ipsae erunt aequales inter se per primam conceptionem: ergo super datam lineam collocavimus triangulum aequilaterum: quod est propositum.
¶ However, if it would be agreeable to assemble the two remaining types of triangles over the same line, namely, a triangle of two equal sides and a triangle of three inequal sides, I will extend the line a.b at each part until the circumferences of both circles will meet upon the two points f and h, and from the center placed at point a, draw the circle e.h.g following the length of the line a.h.	¶ Si autem super eandem lineam libeat collocare reliquas duas triangulorum species, scilicet, triangulum duum aequalium laterum et triangulum trium inaequalium laterum, protrahatur linea .a.b. in utramque partem usquequo occurret circumferentiae amborum circulorum: super duo puncta .f. et .h. et posito centro in puncto .a. lineaetur circulus .e.h.g. secundum quantitatem lineae .a.h.
And likewise, from the center positioned at point b, draw the circle e.f.g following the length of the line b.f.h, and accordingly the circles will intersect each other at two points that are e and g.	Itemque posito centro in puncto .b. lineature circulus .e.f.g. secundum quantitatem lineae .b.f.h. in autem circuli intersecabunt se in duobus punctis qui sunt .e.g.
And so the ends of the given line are joined together with the other of the referred to sections by the two right lines, which are a.g and b.g, and because these lines a.b and a.f exit from the center of the circle c.d.f to its circumference, they will be equal.	Coniungantur igitur extremitates datae lineae cum altera dictarum sectionum per duas lineas rectas quae sunt .a.g.b.g. et quia hae lineae .a.b. et .a.f. exeunt a centro circuli .c.d.f. ad eius circumferentiam ipsae erunt aequales.
As well, because a.b and b.h exit likewise from the center of the circle c.a.d.h to its circumference, they will be equals.	Similiter quoque .a.b. et .b.h. quia exeunt a centro circuli .c.a.d.h. usque ad ipsius circumferentiam ipsae erunt aequales.
Therefore, that both of the lines a.f and b.h are equal to the line a.b, they will be equal among themselves so that by placing a.b in common, b.f will be equal to a.h, but b.f is equal to b.g because both exit from the center of the circle e.f.g to its circumference.	Quia ergo utraque duarum linearum .a.f. et b.h. aequalis est lineae .a.b. ipsae erunt inter se aequales ergo posito .a.b. communi erit .b.f. aequalis .a.h. sed b.f. est aequalis .b.g. quia ambo exeunt a centro circuli .e.f.g. ad eius circumferentiam.
As well, a.h is likewise equal to a.g and both of them are greater than a.b because both of the two lines b.f and a.h are greater than a.b, wherefore we have assembled a triangle of two equal sides upon the given line.	Similiter quoque .a.h. est aequalis .a.g. et utraque earum est maior .a.b. eo quia utraque duarum linearum .b.f. et a.h. maior est .a.b. quare super datam lineam collocavimus triangulum duorum aequalium laterum.
¶ As well, we will place a triangle of three inequal sides upon the same line. If there is any point existing on the circumference to either of the two larger circles that is not within the other of the two sections and which would not obviate f.h, if either side is extended in a continuous and direct manner, then we will connect it by two right lines at both ends of the given line.	¶ Triangulum etiam trium inaequalium laterum super eandem lineam: si aliquod punctum existens in circumferentia alterutrius duorum maiorum circulorum quod non sit in altera duarum sectionum et cui non obviet .f.h. cum in utramlibet partem protracta fuerit in continuum et directum: coniunxerimus per duas lineas rectas cum ambabus extremitatibus datae lineae.
For suppose point k is indicated on the circumference of the circle e.f.g; and it should not be in either section nor obviating f.h when it is extended in a continuous and direct manner up to its circumference. Therefore, I will draw the lines a.k and b.k, and the line a.k will cut the circumference of the circle e.h.g, thus it shall cut at point l and b.k will be equal to a.l because b.k is equal to b.g, and a.l is equal to a.g wherefore a.k is greater than b.k but b.k is greater than a.b, and so the triangle a.b.k is of three inequal sides.	Sit enim punctus .k. signatus in circumferentia circuli .e.f.g. et non sit in altera sectionum nec occurrat ei .f.h. cum protraheretur in continuum et directum usque ad eius circumferentia. protraham ergo lineas .a.k. et b.k. et secabit linea .a.k. circumferentiam circuli .e.h.g. secet ergo in puncto .l. eritque .b.k. aequalis .a.l. quia .b.k. est aequalis .b.g. et .a.l. aequalis .a.g. quare .a.k. est maior .b.k. sed et .b.k. est maior .a.b. triangulus ergo .a.b.k. est trium inaequalium laterum.
We have therefore assembled all types of triangles over the given line.	Sic igitur super datam lineam omnes triangulorum species collocavimus.
Proposition 2	Propositio .2.
[English] Latin	English [Latin][¶]
From a given point, to draw a right line equal to any proposed right line.	A dato puncto cuilibet lineae rectae propositae aequam rectam lineam ducere.
¶ Suppose a is the given point and b.c the given line. From point a, I wish to draw one line equal to the line b.c in whichever direction it may be contiguous, therefore, I will join together point a with one end of the line b.c, whichever I will please; and I will join point a together at the extremity c, by the line a.c, over which I will construct an equilateral triangle according to the doctrine of the preceding, which is to be a.c.d. And at that end of the given line with which I have joined the given point, namely, at the extremity c, I will place the immobile foot of the compass and I will draw over that a circle following the extent of that given line, which is then the circle e.b, and I will lengthen the side of the equilateral triangle that is opposite the given point, namely the side d.c, through the center of the drawn circle and up until its circumference. And suppose the whole line is thus extended according to d.c.e, of whose length I will draw a circle placing the center at d, that is the circle e.f, and I will then draw the side d.a up to the circumference of this last circle, and it shall meet the circumference of that at point f, and so I say that a.f is equal to b.c since b.c and c.e are equal, for that they exit from the center of the circle e.b to its circumference.	¶ Sit .a. punctus datus et .b.c. linea data. volo a puncto .a. ducere lineam unam aequalem lineae .b.c. in quamcumque partem contingatur: coniungam ergo punctum .a. cum altera extremitate lineae .b.c. cum qua voluero. et coniungam ipsum .a. cum extremitate .c. per lineam .a.c. super quam constituam triangulum aequaterum secundum doctrinam praecedentis qui sit .a.c.d. et in illa extremitate lineae datae cum qua coniunxi punctum datum, scilicet, in extremitate .c. ponam pedem circini immobilem et describam super ipsum circulum secundum quantitatem ipsius datae lineae qui sit circulus .e.b. et latus trianguli aequilateri quod opponitur puncto dato scilicet latus .d.c. protraham per centrum circuli descripti usque ad eius circumferentiam: et sit tota linea sic protracta .d.c.e. secundum cuius quantitatem lineabo circulum posito centro in .d. qui sit circulus .e.f. et postea protraham latus .d.a. usque ad circumferentiam huius ultimi circuli et occurrat circumferentiae ipsius in puncto .f. dico igitur quod .a.f. est aequalis .b.c. nam .b.c. et .c.e. sunt aequales quia exeunt a centro circuli .e.b. ad eius circumferentiam.
Similarly, d.f and d.e are also equal because they exit from the center of the circle e.f to the circumference, but d.a and d.c are equal for that they are sides of an equilateral triangle, and so if d.a and d.c are removed from d.e and d.f, which are equals, then the residuals, which are a.f and c.e, are equal. Therefore, because c.e is equal to each of the lines, .a.f. and .c.b., they are equal among themselves, whereby we have extended the line a.f from point a equal to b.c, which is the intent.	Similiter quoque .d.f. et .d.e. sunt aequales quia exeunt a centro circuli .e.f. ad circumferentiam sed .d.a. et .d.c. sunt aequales quia sunt latera trianguli aequilateri. Ergo si .d.a. et .d.c. demantur de .d.e. et .d.f. quae sunt aequales: erunt residua quae sunt .a.f. et .c.e. aequalia. Quia ergo utraque duarum linearum .a.f. et .c.b. est aequalis .c.e. ipsae sunt aequales inter se: quare a puncto .a. protraximus lineam .a.f. aequalem .b.c. quod est propositum.
Proposition 3	Propositio .3.
[English] Latin	English [Latin][¶]
From two proposed inequal lines, to abscind from the longer of them one equal to the shorter.	Propositis duabus lineis inaequalibus de longiori earum breviori aequalem abscindere.
¶ Suppose the two lines are a.b and c.d, and let a.b be the lesser; I wish to abscind from c.d one that will equal a.b, I draw firstly from point c one line equal to a.b following that which the preceding demonstrated, which is c.e, then placing the center at point c, I will draw a circle following the length of c.e that will cut the line c.d, therefore, let it cut so at that point the line c.f will be equal to the line c.e, for that both exit from the center of the same circle to the circumference. And because each of the two lines, a.b and f.c, are equal to c.e, they are equal to one another, which is the intent.	¶ Sint duae lineae .a.b. et .c.d. et sit .a.b. minor volo ex .c.d. abscindere unam quae sit aequalis .a.b. duco primo a puncto .c. unam lineam aequalem .a.b. secundum quod docuit praecedens quae sit .c.e. posito ergo centro in puncto .c. describam circulum secundum quantitatem .c.e. qui secabit lineam .c.d. sit ergo ut secet eam in puncto eritque linea .c.f aequalis lineae .c.e. quia ambo exeunt a centro eiusdem circuli ad circumferentiam: et quia utraque duarum linearum .a.b. et .f.c. est aequalis .c.e. ipsae sunt inter se aequales: quod est propositum.
Proposition 4	Propositio .4.
[English] Latin	English [Latin][¶]
Of all two triangles of which the two sides of one will equal the two sides of the other, and the two angles of them contained by those equilateral sides will one equal the other, then the remaining sides of them will be equal respecting the other and the remaining angles of one will doubtlessly equal the remaining angles of the other, and then even too will the whole triangle equal to the whole triangle.	Omnium duorum triangulorum quorum duo latera unius duobus lateribus alterius aequalia fuerint: duoque anguli eorum illis aequilateribus contenti aequales fuerint alter alteri: latera quoque illorum reliqua sese respicientia aequalia: reliqui vero anguli unius reliquis angulis alterius aequales erunt: ac totus triangulus toti triangulo aequalis.
¶ Suppose there are two triangles, a.b.c and d.e.f, and let the side a.b equal the side d.e, and the side a.c equal the side d.f, and the angle a equal angle d; then I say, that the base b.c is equal to the base e.f and the angle b is equal to angle e.	¶ Sint duo trianguli .a.b.c.d.e.f. sitque latus .a.b. aequale lateri .d.e. et latus .a.c. aequale lateri .d.f. et angulus .a. aequalis angulo .d. tunc dico quod basis .b.c. est aequalis basi .e.f. et angulus .b. aequalis angulo .e.
Further, angle c equals angle f, which is thus demonstrated: I will place the triangle a.b.c over the triangle d.e.f and because angle a falls over angle d and the side a.b over the side d.e, and the side a.c over the side d.f, it is evident by the penultimate conception that neither the angles nor the sides exceed one another, and so because angle a is equal to angle d, then too are the superimposed sides to those which they are superimposed, by the hypothesis, and so points b and c fall upon points e and f. Therefore, if the line b.c falls upon the line e.f, the proposition is evident that when the line b.c is placed over the line e.f, it will neither exceed it nor will it be exceeded by it, for it is equal to it by a conversion of the penultimate conception. And by the same rationale angle b will equal angle e, thus angle c equals angle f, though if the line b.c falls not over the line e.f, but falls within the triangle, just as the line e.g.f or outside like line e.h.f, then two straight lines contain a surface, which is contrary the last petition.	Item angulus .c. aequalis angulo .f. quod probatur: superponam triangulum .a.b.c. triangulo .d.e.f. ita quod angulus .a. cadat super angulum .d. et latus .a.b. super latus .d.e. et latus .a.c. super latus .d.f. et patet per penultimam conceptionem quod nec anguli nec latera sese excedent eo quod angulus .a. est aequalis angulo .d. et latera superposita his quibus superponuntur per ypothesim. puncta ergo .b.c. cadent super puncta .e.f. si ergo linea .b.c. cadit super lineam .e.f. patet propositum: quia cum linea .b.c. superposita lineae .e.f. non excedat eam nec excedatur ab ea est ei aequalis per conversionem penultime conceptionis: eadem ratione erit angulus .b. aequalis angulo .e. et angulus .c. aequalis angulo .f. si autem linea .b.c. non cadit super lineam .e.f. sed cadit inter triangulum sicut linea .e.g.f. aut extra sicut linea .e.h.f. tunc duae lineae recte concludunt superficiem: quod est contra ultimam petitionem.
Proposition 5	Propositio .5.
[English] Latin	English [Latin][¶]
It is necessary for all triangles of two equal sides wherewith the angles above the base are equal, that if its two equal sides are extended in a straight line, then the two angles beneath the base likewise will be made equal.	Omnes trianguli duum aequalium laterum angulos qui supra basim sunt aequales esse necesse est: quod si eius duo aequalia latera directe protrahantur fient quoque sub basi duo anguli invicem aequales.
¶ Let the triangle be a.b.c, whose side a.b is equal to the side a.c. I say that the angle a.b.c is equal to angle a.c.b, because if a.b and a.c are extended up to d and to e, the angle d.b.c will be made equal to the angle e.c.b, which is proven as follows: extending a.b and a.c, I will appoint by the third the line a.d equal to the line a.e, and I will extend the lines e.b and d.c and note the two triangles, a.b.e and a.c.d, and I will demonstrate those to be equal, and equilateral, and equiangular.	¶ Sit triangulus .a.b.c. cuius latus .a.b. sit aequale lateri .a.c. dico quod angulus .a.b.c. est aequalis angulo .a.c.b. quod si protrahantur .a.b. et .a.c. usque ad .d. et .e. fiet angulus .d.b.c. aequalis angulo .e.c.b. quod sic probatur: protractis .a.b. et .a.c. ponam per tertiam lineam .a.d. aequalem lineae .a.e. et protraham lineas .e.b.d.c. et intelligam duos triangulos .a.b.e. et .a.c.d. quos probabo esse aequales et aequilateros et aequiangulos.
So the two sides a.b and a.e of the triangle a.b.c are equal to the two sides a.c and a.d of the triangle a.c.d, and the angle a is common to both, thus by the preceding, the base b.e is equal to the base c.d and the angle e is equal to the angle d and the angle a.b.e is equal to the angle a.c.d.	Sunt enim duo latera .a.b. et .a.e. trianguli .a.b.c. aequalis duobus lateribus .a.c. et .a.d. trianguli .a.c.d. et angulus .a. est communis utrique ergo per praemissam basis .b.e. est aequalis basi .c.d. et angulus .e. est aequalis angulo .d. et angulus .a.b.e. aequalis angulo .a.c.d.
Further note the two triangles d.b.c and e.c.b, for I will demonstrate those likewise to be equilateral and equiangular, for the two sides d.b and d.c of the triangle b.d.c equal the two sides e.c and e.b of the triangle e.b.c, and angle d equals angle e, and so by the preceding the base to the base and the remaining angles to the remaining angles, then the angle d.b.c is thus equal to the angle e.c.b and this is the second intention, namely, that the angles beneath the base are equal. And the angle b.c.d is equal to the angle e.b.c, but the whole of a.b.e is equal to a.c.d, as was proven above, thus the residual angle a.b.c is equal to the residual angle a.c.b, each which is upon the base, which is the first intention.	Item intellige duos triangulos .d.b.c. et .e.c.b. quos similiter probabo esse aequilateros et aequilangulos. nam duo latera .d.b. et .d.c. trianguli .b.d.c. sunt aequalia duobus lateribus .e.c. et .e.b. trianguli .e.b.c. et angulus .d. est aequalis angulo .e. ergo per praemissam basis basi: et reliqui anguli reliquis angulis ergo angulus .d.b.c. est aequalis angulo .e.c.b. et hoc est secundum propositium, scilicet, quod anguli sub basi aequales sunt et angulus .b.c.d. est aequalis angulo .e.b.c. sed totus .a.b.e. est aequalis .a.c.d. ut probatum fuit supra: ergo angulus .a.b.c. residuus est aequalis angulo .a.c.b. residuo quorum uterque est super basim: quod primum propositum.
Proposition 6	Propositio .6.
[English] Latin	English [Latin][¶]
If two angles of any triangle are equal, then the two sides respecting those angles will be equal.	Si duo anguli alicuius trianguli aequales fuerint duoque latera angulos illos respicientia aequalia erunt.
¶ This is a converse to the preceding, as to the first of its part.	¶ Haec est conversa praemissum quantum ad primam eius partem.
For let the triangle be a.b.c of which two angles, b and c, are equal. I say that the side a.b is equal to the side a.c, for if they are not equal, then one or the other will be greater, so let a.b be greater, that it is resected to equality with a.c by the third proposition, and let the superfluous be resected from part a at point d and let b.d equal a.c, and draw the line d.c.	Sit enim triangulus .a.b.c. cuius duo anguli .b. et .c. sunt aequales. dico quod latus .a.b. est aequalis lateri .a.c. Si enim non sunt aequales erit alterum altero maius: sitque .a.b. maius quod resecetur ad aequalitatem .a.c. per tertiam propositionem. ut superfluum sit a parte .a. et resecetur in puncto .d. sitque .b.d. aequalis .a.c. et ducatur linea .d.c.
I thereby note the two triangles, a.b.c and d.b.c, and I will demonstrate those to be equilateral and equiangular, for the two sides d.b and b.c of the triangle d.b.c are equal by the hypothesis, therefore the base d.c is equal to the base b.a and the angle d.c.b equal to the angle a.c.b, namely, a part of the whole, which is impossible.	Intelligo ergo duos triangulos .a.b.c. et d.b.c. quos probabo esse aequilateros et aequiangulos: sunt enim duo latera .d.b. et b.c. trianguli .d.b.c. aequalia per ypothesim ergo basis .d.c. est aequalis basi .b.a. et angulus .d.c.b. aequalis angulo .a.c.b. pars videlicet toti quod est impossibile.
Proposition 7	Propositio .7.
[English] Latin	English [Latin][¶]
If from the two points terminating any line two lines exit concurrently to one point, it is impossible for any other two lines to be drawn from the same points, each one equal to their conterminate, that might concur at any other in the same direction.	Si a duobus punctis aliquam lineam terminantibus duae lineae ad punctum unum concurrentes exierint ab eisdem punctis alias lineas singulas suis conterminalibus aequales quae ad alium concurrant in eandem partem duci est impossibile.
¶ Suppose the line is a.b, from whose ends two lines are extended in one direction, which concur at the same point, as does a.c and b.c, which concur at point c. I say that any other two shall not be extended in the same direction from the same ends that concur at any other point, so let that which will egress from point a be equal to the line a.c, and let that which will egress from point b be equal to the line b.c.	¶ Sit linea .a.b. a cuius extremitatibus protrahantur duae lineae in partem unam quae concurrant in eodem puncto ut sint .a.c. et .b.c. quae concurrant in puncto .c. dico quod in eandem partem non protrahentur aliae duae ab eisdem extremitatibus quae concurrant ad aliud punctum: ita quod illa quae egredietur a puncto .a. sit aequalis lineae .a.c. et quae egredietur a puncto .b. sit aequalis lineae .b.c.
That if it were possible two other lines might be extended in the same direction that concur at point d, then suppose line a.d equals the line a.c and the line b.d equals b.c, and point d will either fall within the triangle or without, for on the other side of a.c and b.c it should fall not, for that a part would then have been equal to its whole.	Quod si fuerit possibile protrahantur aliae duae lineae in eandem partem quae concurrant in puncto .d. et sit linea .a.d. aequalis lineae .a.c. et linea .b.d. aequalis .b.c. aut ergo punctus .d. cadet intra triangulum aut extra: nam in altera laterum .a.c. et .b.c. non cadet quia tunc pars esset aequalis suo toti.
If, however, it falls without, either one of the lines a.d and b.d will cut the other of the lines a.c and b.d, or neither and nor. And the first one shall cut the other and the line c.d is drawn, and that the two sides a.c and a.d of triangle a.c.d are equal, the angle a.c.d will equal the angle a.d.c by means of the fifth.	Si autem cadat extra aut altera linearum .a.d. et .b.d. secabit alteram linearum .a.c. et .b.c. aut neutra neutram. et secet primo altera alteram et protrahatur linea .c.d. quia ergo trianguli .a.c.d. duo latera .a.c. et .a.d. sunt aequalia. erit angulus .a.c.d. aequalis angulo .a.d.c. per quintam.
Similarly, because the two sides b.c and b.d in the triangle b.c.d are equal, the angles b.c.d and b.d.c will be equal.	Similiter quia in triangulo .b.c.d. duo latera .b.c. et .b.d. sunt aequalia erunt anguli .b.c.d. et .b.d.c.
Likewise equal by the same: since the angle b.d.c is greater than the angle a.d.c, it follows for the angle b.c.d to be greater than the angle a.c.d, namely, a part from the whole, which is impossible.	Similiter aequales per eandem: et quia angulus .b.d.c. est maior angulo .a.d.c. sequitur angulum .b.c.d. esse maiorem angulo: a.c.d. partem scilicet toto quod est impossibile.
And if d falls without the triangle a.b.c so that the lines do not intersect, I will extend the line d.c and draw out b.d and b.c beneath the base up to f and to e, and that the line a.d and a.c are equal, the angles a.c.d and a.d.c will be equal by means of the fifth.	Si autem .d. cadit extra triangulum .a.b.c. ita quia lineae non se secent protraham lineam .d.c. et producam .b.d. et b.c. sub basi usque ad .f. et ad .e. et quia lineae .a.d. et .a.c. sunt aequales: erunt anguli .a.c.d. et .a.d.c. aequales per quintam.
Similarly, because b.c and b.d are equal, the angles beneath the base that are c.d.f and d.c.e, will be equal through the second part of the selfsame, and because the angle e.c.d is less than the angle a.c.d then it follows for the angle f.d.c to be less than the angle a.d.c, which is impossible. And in the same way, the adversary will be brought toward inconvenience, if point d were to fall within the triangle a.b.c, and so on.	Similiter quia: .b.c. et .b.d. sunt aequales. erunt anguli sub basi qui sunt .c.d.f. et .d.c.e. aequales per secundam partem eiusdem: quia ergo angulus .e.c.d. minore est angulo .a.c.d. sequitur angulum .f.d.c. esse minorem angulo .a.d.c. quod est impossibile: et eodem modo deducetur adversarius ad inconveniens: si .d. punctus cadat intra triangulum .a.b.c. etc.
Proposition 8	Propositio .8.
[English] Latin	English [Latin][¶]
It is necessary of all two triangles by which the two sides of one will equal the two sides of the other and the base of one will equal the base of the other, for the two angles contained by equal sides to be equal.	Omnium duorum triangulorum quorum duo latera unius duobus lateribus alterius fuerint aequalia: basisque unius basi alterius aequalis: duos angulos aequis contentos aequales esse necesse est:
¶ Suppose the two triangles are a.b.c and d.e.f, and let a.c equal d.f and b.e equal e.f and a.b equal d.e. I say that the angle c is equal to the angle f and the angle a to the angle d and the angle b to the angle e; and that I might place the base a.b over the base d.e, for that they are equal neither shall exceed the other by means of the penultimate conception, and so either point c will fall over point f or it will not.	¶ Sint duo trianguli .a.b.c.d.e.f. sitque .a.c. aequalis .d.f. et .b.e. aequalis: e.f. et .a.b. aequalis .d.e. dico quod angulus .c. est aequalis angulo .f. et angulus .a. angulo .d. et angulus .b. angulo .e. superponam basim .a.b. basi .d.e. quae cum sint aequales neutra excedet alteram per penultimam conceptionem: aut ergo punctus .c. cadet super punctum .f. aut non.
If thus, then because angle c will be placed over angle f and neither of them shall exceed the other, they are equal through a conversion of the aforementioned conception.	Si sic: tunc quia angulus .c. superpositus erit angulo .f. et neuter eorum excedit alterum: ipsi sunt aequales per conversionem conceptionis praedicte.
Likewise show the remaining angles to be equal.	Similiter argue reliquos angulos esse aequales.
But if point c does not fall over f but over any other, then suppose that point is g, because e.g is equal to b.c in the same way, then likewise d.g will equal a.c, e.g will equal e.f, and d.g will equal d.f, which is impossible through the preceding.	Si autem punctus .c. non cadit super .f. sed super quemlibet alium qui sit punctus .g. quia .e.g. est aequalis .b.c. in modo eadem: itemque .d.g. aequalis .a.c. erit .e.g. aequalis .e.f. et .d.g. aequalis .d.f. quod est impossibile per praecedentem.
Proposition 9	Propositio .9.
[English] Latin	English [Latin][¶]
To cut a given angle by equals.	Datum angulum per aequalia secare.
¶ Let the given angle that it behoves to divide be the angle a.b.c. I will place its containing lines that are a.b and b.c, equal by the third, and I will draw the line a.c over which I shall construct the equilateral triangle a.d.c. And I will extend the line b.d, and say that it divides the given line by equals.	¶ Sit datus angulus quem oportet dividere: angulus .a.b.c. lineas ipsum continentes quae sunt .a.b. et .b.c. ponam aequales per tertiam et producam lineam .a.c. super quam constituam triangulum aequalaterum .a.d.c. et protraham lineam .b.d. dico quod ipsa dividit datum angulum per aequalia.
I note the two triangles a.b.d and c.b.d and because the two sides, a.b and b.d, of the triangle a.b.d are equal to the two sides, c.b and b.d, of the triangle c.b.d, and the base a.d to the base c.d, then by the preceding the angle a.b.d is equal to the angle c.b.d, which is to produce the intent.	Intelligo duos triangulos .a.b.d. et .c.b.d. et quia duo latera .a.b. et .b.d. trianguli .a.b.d. sunt aequalia duobus lateribus .c.b. et b.d. trianguli c.b.d. et basis .a.d. basi .c.d. ergo per praecedentem angulus .a.b.d. est aequalis angulo .c.b.d. quod est propositum facere.
Proposition 10	Propositio .10.
[English] Latin	English [Latin][¶]
Proposing a right line to divide it by equals.	Proposita recta linea eam per aequalia dividere.
¶ Let the proposed line that it behoves to divide by equals be the line a.b. Over that I will construct the equilateral triangle a.b.c and divide the angle c by equals, according to doctrine of the preceding by means of the line c.d. I say that the line c.d divides the given line a.b by half.	¶ Sit proposita linea quam oportet dividere per aequalia linea .a.b. super ipsam constituam triangulum aequilaterum .a.b.c. et angulum .c. divido per aequalia secundum doctrinam praecedentis per lineam .c.d. dico quod linea .c.d. dividit datam lineam .a.b. per aequalia.
For I note the two triangles, a.c.d and b.c.d, and thus argue the two sides a.c and c.d of the triangle a.c.d equal the two sides b.c and c.d of the triangle b.c.d, and the angle c of the one to the angle c of the other, and so by means of the fourth, the base a.d equals the base d.b, which is the intent.	Intelligo enim duos triangulos .a.c.d. et .b.c.d. et arguo sic duo latera .a.c. et .c.d. trianguli .a.c.d. sunt aequalia duobus lateribus .b.c. et .c.d. trianguli .b.c.d. et angulus .c. unius angulo .c. alterius ergo per quartam basis .a.d. basi .d.b. quod est propositum.
Proposition 11	Propositio .11.
[English] Latin	English [Latin][¶]
On a given right line, to draw out from a point indicated on it, a perpendicular supported such from both sides by two equal and right angles.	Data linea recta a puncto in ea signato perpendicularem extrahere duobus quidem angulis aequilibus ac rectis utrique subnixam.
¶ Let the given line be a.b whereon is the given point c, from which it is behoven to draw out a perpendicular, and so I will by means of the third make the line b.c equal to the line a.c, and over the whole of the line a.b I constitute the equilateral triangle a.b.d, and extend the line .c.d, of which I say that it is perpendicular to the line a.b.	¶ Sit data linea .a.b. in qua sit datus punctus .c. a quo oportet perpendicularem extrahere. faciam ergo per tertiam lineam .b.c. aequalem lineae .a.c. et super totam .a.b. constituo triangulum aequilaterum .a.b.d. et protraho lineam .c.d. de qua dico quod ipsa est perpendicularis super lineam .a.b.
I note the two triangles a.c.d and b.c.d and for that the two sides a.c and c.d of the triangle a.c.d are equal to the two sides c.b and c.d of the triangle c.b.d, and the base a.d to the base b.d, by means of the 8th the angle a.c.d will equal the angle b.c.d, wherefore each of them will be right by the definition of a right angel. And the line c.b is perpendicular to the line a.b by the definition of a perpendicular line, which is the intent.	Intelligo duos triangulos .a.c.d. et b.c.d. et quia duo latera .a.c. et .c.d. trianguli .a.c.d. sunt aequalia duobus lateribus .c.b. et .c.d. trianguli .c.b.d. et basis .a.d. basi .b.d. erit per 8 angulus .a.c.d. aequalis angulo .b.c.d. quare uterque eorum erit rectus per diffinitionem anguli recti: et linea .c.b. perpendicularis super lineam .a.b. per diffinitionem lineae perpendicularis quod est propositum.
Proposition 12	Propositio .12.
[English] Latin	English [Latin][¶]
From outside a designated point, to draw a perpendicular to a given line of indefinite length.	A puncto extra signato ad datam lineam indefinite quantitatis perpendicularem deducere.
¶ Let the designated point be a, outside the line b.c, from which to that it is behoven to draw a perpendicular. Thus I will extend the line b.c in either directions so much as is pleasing, and over the point a I will draw the circle b.c, so that it may cut the given line at points b and c, and I will extend the lines a.b and a.c, and will divide the angle b.a.c by equals by the line a.d and by means of the 9th, I say that a.d is perpendicular to the line b.c.	¶ Sit .a. punctus signatus extra lineam: b.c. a quo ad ipsam oportet deducere perpendicularem; protraham ergo lineam .b.c. in utramque partem quantum libuerit. et super punctum .a. describam circulum .b.c. sic ut secet lineam datam in punctis .b.c. et protraham lineas .a.b. et .a.c. et dividam angulum .b.a.c. per aequalia per lineam .a.d. per .9. dico quod .a.d. est perpendicularis super lineam .b.c.
I note the two triangles a.b.d and a.c.d and that the two sides a.b and a.d of the triangle a.b.d are equal to the two sides a.c and a.d of the triangle a.c.d, and one angle a is equal to the other angle a, by means of the 4th the base b.d will equal the base d.c and the angle a.d.b will equal the angle a.d.c, whereby both of them are right, and the line a.d is perpendicular to the line b.c by the definition of a right angle, and of a perpendicular line, which is the intent.	Intelligo duos triangulos .a.b.d. et .a.c.d. et quia duo latera .a.b. et .a.d. trianguli .a.b.d. sunt aequalia duobus lateribus .a.c. et .a.d. trianguli .a.c.d. et angulus .a. unius aequalis angulo .a. alterius: erit per 4 basis .b.d. aequalis basi .d.c. et angulus .a.d.b. aequalis angulo .d.b.c. [sic] quare uterque eorum rectus et linea .a.d. perpendicularis super lineam .b.c. per diffinitionem anguli recti et lineae perpendicularis: quod est propositum.
Proposition 13	Propositio .13.
[English] Latin	English [Latin][¶]
Of every right line standing over a right line, the two angles on both sides are either right or equal to two right angles.	Omnis rectae lineae super rectam lineam stantis duo utrobique anguli aut sunt recti aut duobus rectis aequales.
¶ Let the line a.b stand upon the line c.d that, if it is to be perpendicular to it, it will make two right angles through a conversion of the definition. If, however, it were not perpendicular to it, b.e may be drawn perpendicular to c.d from point b by means of the 11th, and the two angles e.b.c and e.b.d will be right through a conversion of the selfsame definition, and that the two angles d.b.a and a.b.e are made equal to the angle d.b.e, it along with angle c.b.e will be equal to two right angles, whereby the three angles, which are d.b.a, a.b.e, and c.b.e, are equal to two right angles. But the angle c.b.a is equal to the two angles c.b.e and e.b.a, therefore, the two angles c.b.a and a.b.d are equal to two right angles, which is the intent.	¶ Sit ut linea .a.b. superstet lineae .c.d. quae si fuerit super eam perpendicularis faciet duos angulos rectos per conversionem diffinitionis. si autem non fuerit super eam perpendicularis a puncto .b. ducatur .b.e. perpendicularis super .c.d. per .11. eruntque duo anguli .e.b.c. et .e.b.d. recti per conversionem diffinitionis: quia ergo duo anguli .d.b.a. et .a.b.e. adaequantur angulo .d.b.e ipse cum angulo .c.b.e. erunt aequales duobus rectis: quare tres anguli qui sunt .d.b.a.a.b.e. et .c.b.e. sunt aequales duobus rectis: sed angulus .c.b.a. est aequalis duobus anguli .c.b.e. et .e.b.a. ergo duo anguli .c.b.a. et .a.b.d. sunt aequales duobus rectis. quod est propositum.
From this it is clear that the total space on the surface of any such plane that surrounds any such point is equal to four right angles.	Ex quo patet quod totum spacium quod in qualibet superficie plana punctum quodlibet circumstat quatuor rectis angulis esse aequale.
Proposition 14	Propositio .14.
[English] Latin	English [Latin][¶]
If two lines exit from a point of one line in two different directions and produce about them right angles or angles equal to two right angles, those two lines are joined directly with one another and are one line.	Si duae lineae a puncto unius lineae in diversas partes exierint duosque circa se angulos rectos aut duobus rectis aequales fecerint: illae duae lineae sibi directae coniunctae sunt et linea una.
¶ Suppose that from point b of the line a.b, two lines exit in opposite directions, which are b.c and b.d, that produce two angles, which are c.b.a and d.b.a, equal to two right angles. Then I say that the lines c.b and d.b are directly joined one to the other and they are one line, and this is like a conversion of the former. That if it were not one line thereupon extended in a direct and continuous manner, since it is not one line with it, d.b will pass over it as b.e or under it as b.f, and so because the line a.b falls over the right line that is c.b.e, the angles c.b.a and e.b.a will be equal to two right angles by the preceding, and because all right angles are reciprocally equal by the 3rd petition and because the angles c.b.a and d.b.a are equal to two right angles by the hypothesis, the two angles c.b.a and e.b.a will equalthe two angles c.b.a and d.b.a. Therefore, by removing the common angle c.b.a, the angle e.b.a will equal the angle d.b.a, a part of the whole, which is impossible. Similarly, the extended line c.b might demonstrate the angle d.b.a is equal to the angle f.b.a, if the adversary were perhaps stating the extended line c.b were to fall below b.d.	¶ Sit ut a puncto .b. lineae .a.b. exeant duae lineae in oppositas partes quae sint .b.c. et .b.d. et faciunt duos angulos qui sint .c.b.a. et .d.b.a. aequales duobus rectis: tunc dico quod duae lineae .c.b. et .d.b. sunt sibi invicem directae coniunctae et linea una et haec est quasi conversa prioris: quod si non fuerit linea una tunc protrahatur .c.b. in continuum et directum quae quia non est linea una cum .d.b. transibit super eam ut .b.e. aut infra eam ut .b.f. quia ergo super lineam rectam qui est .c.b.e. cadit linea .a.b. erunt anguli .c.b.a. et .e.b.a. aequales duobus rectis per precedentem et quia omnes recti sunt adinvicem aequales per .3. petitionem anguli quoque .c.b.a. et .d.b.a. sunt aequales duobus angulus rectis per ypothesim erunt duo anguli .c.b.a. et .e.b.a. aequales duobus angulis .c.b.a. et .d.b.a. ergo dempto communi angulo .c.b.a. erit angulis .e.b.a. aequalis angulo .d.b.a. pars toti: quod est impossibile: similiter lineam .c.b. protractam probabis angulum .d.b.a. esse aequalem angulo .f.b.a. si forte diceret adversarius lineam .c.b. protractam cadere infra .b.d.
Proposition 15	Propositio .15.
[English] Latin	English [Latin][¶]
Of all two lines intersecting one another: all angles positioned opposite themselves are equal, whence it is manifest that when two right lines intersect one another, the four angles that are made are to be equal to four right angles.	Omnium duarum linearum se invicem secantium: omnes anguli contra se positi sunt aequales: unde manifestum est quod cum duae lineae recte se invicem secant quatuor qui fiunt angulos quatuor rectis esse aequales.
¶ Suppose the two lines, a.b and c.d, cut one another at point e. I say that the angle d.e.b is equal to the angle a.e.c and the angle b.e.c is equal to the angle a.e.d, for the two angles a.e.c and c.e.b will, by means of the 13th, be equal to two right angles, and likewise the two angles c.e.b and d.e.b are equal to two right angles by means of the same, wherefore the first two are equal to the last two inasmuch that all right angles are reciprocally equal by the 3rd petition. Thus, by removing the common angle that is c.e.b, the angle a.e.c will be equal to the angle d.e.b.	¶ Sint duae lineae .a.b. et .c.d. se invicem secantes in puncto .e. dico quod angulus .d.e.b. est aequalis angulo .a.e.c. et angulus .b.e.c. est aequalis angulo .a.e.d. erunt enim per .13. duo anguli .a.e.c. et .c.e.b. aequales duobus rectis: itemque duo anguli .c.e.b. et .d.e.b. aequales duobus rectis per eandem: quare duo primi sunt aequales duobus postremis eo quod omnes recti sunt adinuicem aequales per .4. [sic] petitionem: dempto ergo communi angulo quod est .c.e.b. erit angulus .a.e.c. aequalis angulo .d.e.b.
In the same way, the angle c.e.b is demonstrated to be equal to the angle a.e.d, which is the intent.	Eodem modo probabitur angulum .c.e.b. esse aequalem angulo .a.e.d. quod est propositum.
Proposition 16	Propositio .16.
[English] Latin	English [Latin][¶]
If any of the sides of a triangle are directly extended, the angle produced from without will be greater than either angle opposite itself from within the triangle.	Si quotlibet laterum trianguli directe protrahatur faciet angulum extrinsecum utroque angulo trianguli sibi intrinsecus opposito maiorem.
¶ Suppose that side a.b. of the triangle a.b.c is extended up until d. I say that the angle d.b.c is greater than either of the two angles from within and opposite itself, which are b.a.c and b.c.a, for I will divide by means of the 10th the line c.b by equals at point e, and I will extend a.e up until f so e.f shall be made equal to a.e, and I will extend the line f.b. And I note the two triangles, c.e.a and b.e.f, and for that the two sides a.e and e.c of the triangle a.e.c are equal to the two sides f.e and e.b of the triangle f.e.b, and that the angle e of one is equal to the angle e of the other by the preceding, for the angles are placed opposite, then by the 4th the angle e.c.a will equal the angle e.b.f, and the angle e.b.d will be greater than angle the b.c.a.	¶ Sit ut trianguli .a.b.c. latus .a.b. protrahatur usque ad .d. dico quod angulus .d.b.c. maior est utroque duorum angulorum intrinsecorum sibi oppositorum qui sunt .b.a.c. et .b.c.a. dividam enim per .10. lineam .c.b. per aequalia in puncto .e. et protraham .a.e. usque ad .f. ita ut .e.f. fiat aequalis .a.e. et protraham lineam .f.b. intelligo duos triangulos .c.e.a. et .b.e.f. et quia duo latera .a.e. et .e.c. trianguli .a.e.c. sunt aequalia duobus lateribus .f.e. et .e.b. trianguli .f.e.b. et angulus .e. unius est aequalis angulo .e. alterius per praemissam: quia sunt anguli contra positi: erit per .4. angulus .e.c.a. aequalis angulo .e.b.f. et ideo angulus .e.b.d. maior erit angulo .b.c.a.
It likewise will be similarly proved that it is greater than the angle c.a.b, for I will divide a.b by equals at point g by the 10th and extend the line g.h equal to the line c.g by the 3rd. Then I will extend h.b.k, and of the two triangles that are a.g.c and b.g.h, the two sides a.g and g.c of the first will be equal to the two sides b.g and b.h of the second, and the angle g of one to the angle g of the other by the 15th, therefore, by the 4th the angle g.a.c is equal to the angle g.b.h and so by the 15th to the angle k.b.d. And because the angle c.b.d is greater than the angle k.b.d, it will also be greater than the angle b.a.c, which is the intent.	Similiter quoque probabitur quod est maior angulo .c.a.b. nam dividam .a.b. per aequalia in puncto .g. per .10. et protraham lineam .g.h. aequalem lineae .c.g. per .3. postea protraham .h.b.k. eruntque duorum triangulorum qui sunt .a.g.c. et .b.g.h. duo latera .a.g. et .g.c. primi aequalia duobus lateribus .b.g. et .b.h. secundi: et angulus .g. unius angulo .g. alterius per .15. ergo per .4. angulus .b.c.a. [sic] est aequalis angulo .g.b.h. quare per .15. et angulo .k.b.d. et quia angulus .c.b.d. est maior angulo .k.b.d. erit etiam maior angulo .b.a.c. quod est propositum.
Proposition 17	Propositio .17.
[English] Latin	English [Latin][¶]
Any two angles of every triangle are less than two right angles.	Omnis trianguli duo quilibet anguli duobus rectis sunt minores.
¶ Suppose the triangle is a.b.c. I say that any two of its angles are smaller than two right angles, for that one side of it is extended, just as b.c up until d, by the preceding the angle c from without will be greater than a and greater than b. But c from without with c from within is equal to two right angles by means of the 13th, thus angles b and c from within or angles a and c from within are less than two right angles, and similarly, if the side b.a is extended, it will be proved that the two angles a and b are smaller than two right angles, which is the intent.	¶ Sit triangulus .a.b.c. dico quod duo quilibet eius anguli duobus rectis sunt minores. protrahatur enim unum latus eius ut .b.c. usque ad .d. eritque per praecedentem angulus .c. extrinsecus maior .a. et maior .b. sed .c. extrinsecus cum .c. intrinseco est aequalis duobus rectis per .13. ergo anguli .b. et .c. intrinseci sive anguli .a. et .c. intrinseci sunt minores duobus rectis: similiter si protrahatur latus .b.a. probabitur quod duo anguli .a. et .b. sunt minores duobus rectis quod est propositum.
Proposition 18	Propositio .18.
[English] Latin	English [Latin][¶]
The longest side of every triangle is opposite its greatest angle.	Omnis trianguli longius latus maiori angulo oppositum est:
¶ Suppose that on the triangle a.b.c, the angle a is greater than the angle c. I say that the side c.b will be greater than the side a.b, for if it were equal by means of the 5th, angle a would equal angle c, which is contrary the hypothesis. If, however, a.b is greater, then it may be resected to equality with c.b by means of the 3rd and d.b shall equal c.b. Therefore, by means of the 5th, the angle d.c.b will be equal the angle b.d.c, but b.d.c is greater than the angle b.a.c by means of the 16th, and so b.c.d is greater than b.a.c, thus even more greater than a.c.b, a part from the whole, which is impossible.	¶ Sit ut in triangulo .a.b.c. angulis .a. sit maior angulo .c. dico quod latus .c.b. maior erit latere .a.b. si enim sit aequale erit per .5. angulis .a. aequalis angulo .c. quod est contra ypothesim: si autem .a.b. sit maius resecetur ad aequalitatem .c.b. per .3. sitque .d.b. aequale .c.b. erit ergo per .5. angulus .d.c.b. aequalis angulo .b.d.c. sed .b.d.c. est maior angulo .b.a.c. per .16. ergo .b.c.d. est maior .b.a.c. quare multo fortius maior .a.c.b. pars toto: quod est impossibile.
Proposition 19	Propositio .19.
[English] Latin	English [Latin][¶]
The greatest angle of every triangle is opposite its longest side.	Omnis trianguli maiori angulo longius latus oppositum est.
¶ Suppose that on the triangle a.b.c, the side b.c is greater than the side a.b. I say that the angle a will be greater than the angle c, and is converse the preceding, for if it were equal then by means of the 6th then the side a.b would equal the side b.c, which is contrary the hypothesis. If, however, c is greater, then by the preceding the side a.b is greater than the side b.c, which is contrary the hypothesis, whereby the proposition is construed.	¶ Sit ut in triangulo .a.b.c. latus .b.c. sit maius latere .a.b. dico quod angulus .a. erit maior angulo .c. et est conversa praecedentis: si enim sit aequalis tunc per .6. latus .a.b. est aequale lateri .b.c. quod est contra hypothesim. si autem .c. sit maior tunc per praecedentem latus .a.b. est maius latere .b.c. quod est contra ypothesim quare astruitur propositum.
Proposition 20	Propositio .20.
[English] Latin	English [Latin][¶]
Any two sides of every triangle joined together are longer than the remaining.	Omnis trianguli duo quaelibet latera simul iuncta reliquo sunt longiora.
¶ Suppose the triangle is a.b.c. I say that the two sides, a.b plus a.c, are longer than the side b.c, and were the line b.a extended up until d so that a.d is equal to a.c, and c.d extended, then by the fifth the angle a.c.d will be equal to the angle d, whereby the angle b.c.d is greater than angle d. Therefore, by the 18th the side b.d is greater than the side b.c, but b.d is equal to a.b plus a.c and thus b.a and a.c joined together are greater than b.c.	¶ Sit triangulus .a.b.c. dico quod duo latera .a.b. et a.c sunt longiora latere .b.c. protrahatur linea .b.a. usque ad .d. ita ut .a.d. sit aequalis .a.c. et protrahatur .c.d per quintam erit angulus .a.c.d. aequalis angulo .d. quare angulus .b.c.d. est maior angulo .d. ergo per .18. latus .b.d. est maius latere .b.c. sed .b.d. est aequale .a.b. et .a.c. quare .b.a. et .a.c. simul iuncta sunt maiora .b.c.
Proposition 21	Propositio .21.
[English] Latin	English [Latin][¶]
If two lines exiting from the two terminal points of one side of a triangle convene at one point within the triangle itself, then likewise the two remaining lines of the triangle will be shorter and shall contain the greater angle.	Si de duobus punctis terminalibus unius lateris trianguli duae lineae exeuntes intra triangulum ipsum ad punctum unum conueniant eadem duabus quidem reliquis trianguli lineis breviores erunt et maiorem angulum continebunt.
¶ Suppose that from the extremities of the side b.c on the triangle a.b.c, the two lines b.d and c.d shall concur at point d within the triangle a.b.c. I say that those joined together are shorter than the two lines a.b and a.c joined together, and that the angle d is greater than angle a, for I will extend b.d until whereto it shall cut the side a.c at point e, and by the 20th b.a and a.e joined together will be greater than b.e, thus b.a and a.c are greater than b.e and e.c.	¶ Sit ut in triangulo .a.b.c. ab extremitatibus laterius .b.c. concurrant duae lineae .b.d. et .c.d. ad punctum .d. intra triangulum .a.b.c. dico quod ipsae simul iunctae sunt breviores duabus lineis .a.b. et .a.c. simul iunctis et quod angulus .d. est maior angulo .a. protraham enim .b.d. usque quo secet latus .a.c. in puncto .e. eruntque per .20. .b.a. et a.e. simul iunctae maiores .b.e. ergo .b.a. et .a.c. sunt maiores .b.e. et .e.c.
And moreover, d.e and e.c joined together by the same means are greater than d.c, whereby b.c and e.c are greater than b.d and d.c. And because b.a and a.c are greater than b.e and e.c, as was proved before, they will be even more greater than b.d and d.c, which is the first intention.	At vero .d.e. et .e.c. simul iunctae per eandem sunt maiores .d.c. quare .b.c. et .e.c. sunt maiores .b.d. et .d.c. et quia .b.a. et .a.c. sunt maiores .b.e. et .e.c. ut probatum est prius: erunt multo fortius maiores .b.d. et .d.c. quod est primum propositum.
And since the angle b.d.c is greater than the angle d.e.c by the 16th, and the angle d.e.c is greater than the angle e.a.b by the same, the angle b.d.c will be even more greater than the angle b.a.c, which is the second intention.	At quoniam angulus .b.d.e. est maior angulo .d.e.c. per .16. et angulus .d.e.c. est maior angulo .e.a.b. per eandem erit angulus .b.d.c multo fortius maior angulo b.a.c. quod est secundum propositum.
Proposition 22	Propositio .22.
[English] Latin	English [Latin][¶]
To propose three right lines, any two of which joined together are longer than the remaining, by which to construct a triangle from three other lines equal to them.	Propositis tribus lineis rectis quarum duae quaelibet simul iunctae reliquae sint longiores de tribus aliis lineis sibi equalibus triangulum constituere.
¶ Let the three right lines proposed be a, b, and c, and that any two joined together are longer than the remaining, for otherwise a triangle could not be constructed from those equal to the three by means of the 20th. And when I wish to construct a triangle from those three aforesaid lines, I then take the right line, which is d.e—to which I do not appoint a definite end at part e—from which I establish by means of the 3rd d.f equal to a, and f.g equal to b, and g.h equal to c. And I make point f a center and draw out the circle d.k following the length of the line f.d, and likewise, I make g.a a center and draw out the circle k.h following the length of the line g.h. And the circles will intersect each other at two points, one of which will be k; else it would follow that one of the stated lines is to be equal to the other two joined — or greater to them—which is contrary to the position. And so I draw the line k.f and k.g and the triangle k.f.g will be constructed from the three lines equal to the given lines (a, b, and c), for f.d and f.k are equal since they are from center to circumference whereby f.k is equal to a, and similarly, g.h and g.k are equal because they exit from the center to the circumference whereby g.k is equal to c. And because g.f has been assumed to be equal to b, the proposition is clearly evident.	¶ Sint tres lineae rectae propositae .a.b.c. et sint quaelibet duae simul iuncte longiores reliqua: aliter enim ex illis tribus aequalibus triangulus non posset constitui per .20. cum ergo ex illis tribus praedictis volo constituere triangulum: summo lineam rectam quae sit .d.e. cui non pono a parte .e. determinatum finem: de qua sumo per .3. d.f. aequalem .a. et .f.g. aequalem .b. et .g.h. aequaelem .c. factoque puncto .f. centro describo secundum quantitatem lineae .f.d. circulum .d.k. itemque facto .g. centro describo secundum quantitatem lineae .g.h. circulum .k.h. qui circuli intersecabunt se in duobus punctis quorum unum sit .k. alioquin sequeretur unam dictarum linearum esse aequalem aliis duabus iunctis aut maiorem eis: quod est contrarium positioni: duco ergo lineam .k.f. et .k.g. eritque triangulus .k.f.g. constitutus ex tribus lineis aequalibus lineis .a.b.c. datis: sunt enim .f.d. et .f.k. aequales quoniam sunt a centro ad circumferentiam quare .f.k. est aequalis .a. similiterque .g.h. et .g.k. sunt aequales: quia exeunt a centro ad circumferentiam: quare .g.k. est aequalis .c. et quia .g.f. sumpta fuit aequalis .b. patet propositum manifeste.
Proposition 23	Propositio .23.
[English] Latin	English [Latin][¶]
To design an angle equal to any proposed angle over the terminus of a given right line.	Data recta linea super terminum eius cuilibet angulo proposito aequum angulum designare.
¶ Let the given line be f.e, which is on the preceding figure, and the lines b and a shall contain the given angle to which I will subtend the base c. Over point f on the line e.f I shall bid to make an angle equal to the given angle. To the line e.f I am to join f.d, equal to line a, and select f.g from f.e equal to b, and select g.h from g.e equal to c, and over point f and g I draw out two circles, d.k and k.h, following the length of the two lines f.d and g.h, and they intersect each other at point k, just as the preceding demonstrated, and in so drawing the lines k.f and k.g, the two sides k.f and f.g of the triangle k.f.g will be equal to the two sides a and b of the triangle a.b.c, and the base g.k will be equal to the base c. Therefore, by the eighth the angle k.f.g shall equal the angle contained at a and b, which is the intent.	¶ Sit data linea .f.e. quae est in superiori figura: et sint lineae .b.a. continentes angulum datum cui subtendam basim .c. super punctum .f. lineae .e.f. iuberem facere aequalem angulum angulo dato ad lineam .e.f. adiungo .f.d. aequalem lineae .a. et ex .f.e. summo .f.g. aequalem .b. et ex .g.e. summo .g.h. aequalem .c. et super puncta .f. et .g. describo duos circulos .d.k. et k.h. secundum quantitatem duarum linearum .f.d. et .g.h. et intersecantes se in puncto .k. sicut docuit praecedens ductisque lineis .k.f. et .k.g. erunt aequalia duoo latera .k.i. [sic] et .f.g. trianguli .k.f.g. duobus lateribus .a. et .b. trianguli .a.b.c. et basis .g.k. aequalis basi .c. ergo per octavam angulus .k.f.g. aequalis erit angulo contendo ab .a. et .a.b. [sic] quod est propositum.
Proposition 24	Propositio .24.
[English] Latin	English [Latin][¶]
Of all two triangles among which the two sides of one shall equal the two sides of the other, if one of the angles contained by those equal sides will be greater than the other, the base of the same will likewise be greater than the base of the other.	Omnium duorum triangulorum quorum duo latera unius duobus lateribus alterius fuerint aequalia: si fuerit angulorum sub illis aequis lateribus contentorum alter altero maior basis quoque eiusdem basi alterus maior erit.
¶ Suppose the two triangles are a.b.c and d.e.f and let the two sides a.b and a.c equal the two sides d.e and d.f, each to its correlative; the right, namely, to the right, and the left to the left, and let the angle a be greater than the given angle d. I say that the base b.c will be greater than the base e.f, so following the doctrine of the preceding I will make e.d.g equal to angle a and the angle e.d.f will be a part of the angle e.d.g, and I will make d.g equal to a.c and extend e.g, which will either traverse above e.f, as to cut the line d.f, or over e.f, as to be one line with it, or below.	¶ Sint duo trianguli .a.b.c.d.e.f. sintque duo latera .a.b. et .a.c. aequalia duobus lateriubus .d.e. et .d.f. et unumquodque suo correlativo dextrum scilicet dextro: sinistraumque sinistro: sitque angulus .a. maior angulo .d. dato: dico quod basis .b.c. maior erit basi .e.f. faciam enim iuxta doctrinam praecedentis .e.d.g. aequalem angulo .a. eritque angulus .e.d.f. pars eius et ponam .d.g. aequalem .a.c. et protraham .e.g. quae aut transibit supra .e.f. ut secet lineam .d.f. aut super .e.f. ut sit secum linea una: aut infra.
First then, let it traverse above, and because the sides a.b and a.c of the triangle a.b.c are equal to the sides e.d and d.g of the triangle e.d.g, and the angle a is the total of the angle d, by the 8th the base b.c will equal the base e.g. Moreover, because d.g and d.f are equal, for each is equal to a.c, then by the 5th the angle d.f.g equals the angle d.g.f whereby d.f.g will be greater than f.g.e, and so e.f.g is even more greater than the selfsame f.g.e, and so by the 18th the side e.g is greater than the side e.f and so b.c is greater than e.f, which is the intent.	Transeat ergo primo supra et quia .a.b. et .a.c. latera trianguli .a.b.c. sunt aequalia .e.d. et .d.g. lateribus trianguli .e.d.g. et angulus .a. angulo .d. totali: erit per .8. basis .b.c. aequalis basi .e.g. at vero quia .d.g. et .d.f. sunt aequales nam utraque est aequalis .a.c. erit per .5. angulus .d.f.g. aequalis angulo .d.g.f. quare .d.f.g. maior erit .f.g.e. ergo .e.f.g. multo fortius maior est eodem .f.g.e. ergo per .18. latus .e.g: maius est latere .e.f. quare et .b.c. maior est .e.f. quod est propositum.
But if e.g were to traverse over e.f and so should be one line with it, then e.f will be part of e.g, and by means of the final conception the proposition is evident. And if e.g. were to traverse below e.f, the two lines d.f and d.g, which are equal as was demonstrated, are extended up until k and to h, and by the second part of the fifth the angles k.f.g and f.g.h beneath the base f.g will be made equal, whereas the angle e.f.g will be greater than the angle f.g.e, and so by the 18th the side e.g is greater than the side e.f, thus b.c is greater than e.f, which is the intent.	Si vero .e.g. transeat super .e.f. et sit secum lineam una tunc .e.f. erit pars .e.g. per ultimam ergo conceptionem patet propositum. si vero .e.g. transeat infra .e.f. protrahantur duae lineae .d.f. et .d.g. quae sunt aequales ut probatum est usque ad .k. et ad .h. fientque per secundum partem quinti sub basi .f.g. anguli .k.f.g. et .f.g.h. aequales quare angulus .e.f.g. maior erit anglo .f.g.e. ergo per .18. latus .e.g. maius est latere .e.f. quare b.c. maior est e.f. quod est propositum.
That final part may also be demonstrated through 21, for in that third disposition the two lines d.g and e.g are greater than the two lines d.f and f.e, and because d.g is equal to d.f on the account that both are equal to a.c, e.g will be greater than e.f. Thus b.c will be greater than e.f, which is the intent (though it is preferable to demonstrate in the way previous, as every disposition therein is proved through the fifth).	Istud ultimum membrum posset etiam probari per .21. per ipsam enim erunt in dispositione tertia duae lineae .d.g. et .e.g. maiores duabus lineis .d.f. et .f.e. et quia: .d.g. est aequalis .d.f. propter hoc quod ambe sunt aequales .a.c. erit .g.e. maior .e.f. quare et .b.c. maior quod est propositum: melius tamen est demonstrare priori modo ut in omni dispositione arguatur per quintam.
Proposition 25	Propositio .25.
[English] Latin	English [Latin][¶]
Of all two triangles among which the two sides of one shall equal the two sides of the other, and yet the base of one shall be greater than the base of the other: so too will the angle contained within those equal sides of the greater triangle be respectfully greater than the other angle.	Omnium duorum triangulorum quorum duo latera unius duobus lateribus alterius fuerint aequalia: basis vero unius basi alterius fuerit maior erit quoque angulus trianguli maioris illis aequis lateribus contentus angulo alterius se respiciente maior.
¶ Suppose the two triangles are a.b.c and d.e.f and let the two sides a.b and a.c of the first equal the two sides d.e and d.f of the second, each to its correlative, and suppose the base b.c is greater than the base e.f. I say that angle a is greater than angle d, and this is converse the preceding. Indeed, it will not be equal, for then it follows by the 4th that were the base b.c equal to the base e.f, that is contrary the hypothesis, but neither would it be slighter, for it follows then that d would be greater. And so by the preceding the base e.f will be greater than the base b.c, which is contrary to the proposition whereby it will be greater, and in this way the proposition is construed.	¶ Sint duo trianguli .a.b.c.d.e.f. sintque duo latera .a.b. et .a.c. primi aequalia duobus lateribus .d.e. et .d.f. secundi unumquodque suo correlatio: sitque basis .b.c. maior basi .e.f. dico quod angulis .a. maior erit angulo .d. haec est conversa praecedentis. Aequalis quidem non erit: sic enim esset per .4. basis .b.c. aequalis basi .e.f. quod est contra ypothesim: sed nec minor quia sic esset .d. maior: et ita per praecedentem basis .e.f. erit maior basi .b.c. quod est contrarium propositioni quare maior erit sicque propositum astruitur.
Proposition 26	Propositio .26.
[English] Latin	English [Latin][¶]
Of all two triangles among which the two angles of the one, each with respect to the other, shall equal the two angles of the other, and a side of the one shall equal a side of the other: if that side should either be between those two equal angles or opposite one of them, then the two remaining sides of the one, each with respect to the other, will likewise equal the two remaining sides of the other triangle, and the remaining angle of the one will equal the remaining angle of the other.	Omnium duorum triangulorum quorum duo anguli unius duobus angulis alterius et uterque se respicienti aequales fuerint latus quoque unius lateri alterius aequale: fueritque latus illud inter duos angulos aequales aut uni eorum oppositum erunt quoque duo unius reliqua latera duobus reliquis alterius trianguli lateribus unumquodque se respicienti aequalia: angulusque reliquus unius angulo reliquo alterius aequalis.
¶ Suppose the two triangles are a.b.c and d.e.f and let angle b equal angle e and angle c equal angle f and let the side b.c equal the side e.f, or suppose the first two sides a.b and a.c are equal to the other two sides d.e and d.f, so that a.b is equal to d.e or a.c to d.f. I say that the remaining two sides of one will equal the remaining two sides of the other, and the remaining angle will equal the remaining angle; namely, the angle a to the angle d.	¶ Sint duo trianguli .a.b.c.d.e.f. sitque angulus .b. aequalis angulo .e. et angulus .c. aequalis angulo .f. sitque latus .b.c. aequale lateri .e.f. aut alterum duorum laterum .a.b. et .a.c. aequale alteri duorum laterum .d.e. et .d.f. ita quod .a.b. sit aequale .d.e. aut .a.c.d.f. dico quod reliqua duo latera unius erunt aequalia: reliquis duobus lateriubs alterius et reliquus angulus reliquo angulo aequalis: angulus videlicet .a. angulo .d.
And so first I will make the side b.c, on which the angles b and c lie equal to the side e.f, on which the angles e and f lie, which are positioned equal to the angles b and c. Then I say that the side a.b is equal to the side d.e and the side a.c to the side d.f and the angle a to the angle d.	Ponam ergo primo ut latus .b.c. super quod iacent anguli .b.c. sit aequale lateri .e.f. super quod iacent anguli .e.f. qui positi sunt aequales angulis .b.c. tunc dico quod latus .a.b. est aequale lateri .d.e. et latus .a.c. lateri .d.f. et angulus .a. angulo .d.
For if the side a.b shall not equal the side d.e, one will be greater, therefore, suppose d.e is greater, which I will resect to equality with a.b, and let g.e equal a.b. And I will extend the line g.f and by means of the 4th, the angle g.f.e will equal the angle a.c.b and thus the angle d.f.e, a part of the whole, which is impossible. Accordingly, d.e will equal a.b and so by the 4th, d.f will equal a.c and angle d will equal angle a, which is the first part of the divided proposition.	Si enim latus .a.b. non sit aequale lateri .d.e. alterum erit maius: sit ergo maius .d.e. quod resecabo ad aequalitatem .a.b. sitque .g.e. aequale .a.b. et producam lineam .g.f. eritque per .4. angulus .g.f.e. aequalis angulo .a.c.b. quare et angulo .d.f.e. pars toti quod est impossibile: erit ergo .d.e. aequale .a.b. ergo per .4. .d.f. aequale .a.c. et angulus .d. aequalis angulo .a. quod est primum membrum divisionis proposite.
Let the two angles b and c return as before, equal to the two angles e and f, and let the side a.b, which is opposite the angle c, equal the side d.e, which is opposite the angle f, whereby it is positioned equal to the angle c. I say that the side b.c will be equal to the side e.f and the side a.c to the side d.f and angle a to angle d. For if the side e.f will not be equal to the side b.c then one will be greater, therefore suppose e.f is greater, and so e.g is made equal to b.c and I will extend the line d.g, and by the 4th the angle d.g.e will be equal to the angle a.c.b and thus to d.f.e, namely, that from without will equal that from within, which is impossible by the 16th. e.f will thus be equal to b.c by the 4th, and so the side d.f will equal the side a.c and angle d will total angle a, which is the second part of the divided proposition, wherefore the whole is clearly evident.	Sint rursus ut prius duo anguli .b. et .c. aequales duobus angulis .e. et .f. sitque latus .a.b. quod opponitur angulo .c. aequale lateri .d.e. quod opponitur angulo .f. cui positus est aequalis angulus .c. dico quod latus .b.c. erit aequale lateri .e.f. et latus .a.c. lateri .d.f. et angulus .a. angulo .d. si enim latus .e.f. non fuerit aequale lateri .b.c. erit alterum maius: sit ergo .e.f. maius: ponatur itaque .e.g. aequale .b.c. et producam lineam .d.g eritque per .4. angulus .d.g.e. aequalis angulo .a.c.b. quare et angulo .d.f.e. extrinsecus videlicet intrinseco quod est impossibile: per .16. erit ergo .e.f. aequale .b.c. ergo per .4. latus .d.f. aequale lateri .a.c. et angulus .d. totalis angulo .a. quod est secundum membrum divisionis proposite: quare totum manifeste patet.
Proposition 27	Propositio .27.
[English] Latin	English [Latin][¶]
If a right line falls on two right lines and the two coalternate angles shall be reciprocally made equal to one another, those two lines will be equidistant.	Si recta linea super duas lineas rectas ceciderit duosque angulos coalternos sibi invicem aequales fecerit illae duae lineae erunt aequidistantes.
¶ Let the line a.b fall on the two lines c.d and e.f, and cut the line c.d at point g and the line e.f at point h, and suppose the angle d.g.h equals the angle e.h.g. I say that the lines c.d and e.f are equidistant.	¶ Sit ut linea .a.b. cadat super duas lineas .c.d. et .e.f. et secet lineas .c.d. in puncto .g. et lineam .e.f. in puncto .h. sitque angulus .d.g.h. aequalis angulo .e.h.g. dico quod lineae .c.d. et .e.f. sunt aequidistantes.
For if they are not concurrent, either c and e would converge at point k or d and f at point l, and just howsoever it may fall would be impossible by the 16th, namely, for an angle from without to be equal to an angle from within, thus one of the stated coalternate angles of whose placements are equal will be from without and the remaining will be from within, and so since it is impossible that they should concur if extended in either direction, then by the very definition they will be equidistant, which is the intent.	Si enim non concurrant aut ad partem .c.e. super punctum .k. aut a parte .d.f. super punctum .l. et qualitercumque fuerit accidet impossibile per .16. videlicet angulum extrinsecum esse aequalem intrinseco: nam unus dictorum angulorum coalternorum qui positi sunt aequales erit extrinsecus et reliquus intrinsecus quia igitur impossibile est eas concurrere in alterutram partem protractas ipsae per diffinitionem erunt aequidistantes: quod est propositum.
Proposition 28	Propositio .28.
[English] Latin	English [Latin][¶]
If a right line should fall upon two right lines and an angle from within equals the angle opposite itself from without, or if two interior angles taken from one side equal two right angles, those two lines will be equidistant.	Si linea recta duabus lineis rectis supervenerit fueritque angulus eius intrinsecus angulo extrinseco sibi opposito aequalis aut duo anguli intrinseci ex una parte duobus angulis rectis aequales illae duae lineae aequidistantes erunt.
¶ Suppose the line a.b cuts the two lines, c.d and e.f at point g and h, and let the angle g from without equal the angle h, taken from the same side from within, or that the two angles g and h taken from within and from the same side will equal two right angles. I say then that the two lines c.d and e.f are equidistant.	¶ Sit ut linea .a.b. secet duas lineas .c.d. et .e.f. in puncto .g. et .h. sitque angulus .g. extrinsecus aequalis angulo .h. intrinseco ex eadem parte sumpto: aut duo anguli .g. et .h. intrinseci ex eadem parte sumpti sint aequales duobus angulis rectis: dico quod duae lineae .c.d. et .e.f. sunt aequidistantes.
Therefore, first suppose the angle d.g.a equals the angle f.h.g and by the 15th, the angle c.g.h will equal that selfsame angle f.h.,g whereby through the preceding, c.d and e.f are equidistant.	Sit ergo primo angulus .d.g.a. aequalis angulo .f.h.g. eritque per .15. angulus .c.g.h. aequalis eidem angulo .f.h.g. quare per praemissam .d. [sic] et .e.f. sunt aequidistantes.
In turn, let the two angles d.g.h and f.h.g equal two right angles, and that through the 13th the two angles d.g.h and c.g.h are also equal to two right angles, the angle c.g.h will equal the angle f.h.g and thus, by the preceding, c.d and e.f will be equidistant, which is the intent.	Sint rursus duo anguli .d.g.h. et .f.h.g. aequales duobus rectis: et quia per .13. duo anguli .d.g.h. et .c.g.h. sunt similiter aequales duobus rectis erit angulus .c.g.h. aequalis angulo .f.h.g. quare per praemissam .c.d. et .e.f. erunt aequidistantes: quod est propositum.
Proposition 29	Propositio .29.
[English] Latin	English [Latin][¶]
If a line should fall upon two equidistant lines, the two coalternate angles will be equal, and the angle from without will equal the angle opposite it from within. Likewise, two the interior angles from either constituent part shall be equal to two right angles.	Si duabus lineis aequidistantibus linea supervenerit duo anguli coalterni aequales erunt: angulusque extrinsecus angulo intrinseco sibi opposito aequalis. Itemque duo anguli intrinseci ex alterutra parte constituti duobus rectis angulis aequales.
¶ Suppose the two lines a.b and c.d are equidistant, over which the line e.f falls, dividing them at points g and h. I say that the coalternate angles g and h are equal, and that the angle g from without is equal to the angle h from within, each opposite the other, taken from the same side, and that angle g and h taken from the same side are equal to two right angles, and this is converse the two preceding.	¶ Sint duae lineae .a.b. et .c.d. aequidistantes super quas cadat linea .e.f. secans eas in punctis .g. et .h. dico quod anguli .g. et .h. coalterni sunt aequales: et quod angulus .g. extrinsecus est aequalis angulo .h. intrinseco sibi opposito ex eadem parte sumpto: et quod anguli .g. et .h. intrinseci ex eadem parte sumpti sunt aequales duobus rectis. et haec est conversa duarum praecedentium.
The first is thus evident:	Primum sic patet.
that if the angle b.g.h is not equal to the angle c.h.g, the other of them will be greater. Suppose then the angle c.h.g is greater, and because the two angles c.h.g and g.h.d are equal to two right angles, and by the 13th the two angles b.g.h and d.h.g are less than two right angles, thus by the fourth petition, if the two lines a.b and c.d are extended they will meet at part b and d and at some point, such as at k. Therefore, they are not equidistant by definition, which is contrary the hypothesis, and because this is impossible, then consequently the two coalternate angles b.g.h and c.h.g will be equal, which is the first intention. From this the second is evident, for by the 15th the angle b.g.h is equal to the angle a.g.e, and so the angle a.g.e will equal the angle c.h.g, namely the exterior to the interior, which is the second intention.	Si enim angulus .b.g.h. non est aequalis angulo .c.h.g alter eorum erit maior: sit ergo maior angulus .c.h.g. et quia duo anguli .c.h.g. et .g.h.d. sunt aequales duobus rectis ergo per .13. erunt duo anguli .b.g.h. et .d.h.g. minores duobus rectis ergo per quartam duae lineae .a.b: et .c.d. si protrahantur concurrent in parte .b. et .d. ad punctum aliquem ut ad .k. non ergo sunt aequidistantes per diffinitionem quod est contra ypothesim: et quia hoc est impossibile: erunt igitur duo anguli coalterni .b.g.h. et .c.h.g. aequales quod est primum propositum. Ex hoc patet secundum: est enim per .15. angulus .b.g.h. aequalis angulo .a.g.e. ergo angulus .a.g.e. erit aequalis angulo .c.h.g. extrinsecus videlicet intrinseco: quod est secundum propositum.
From this again the third is evident. For by the 13th the two angles a.g.e and a.g.h are equal to two right angles, and so the two angles a.g.h and c.h.g that are two from within, taken from the same side, are also equal to two right angles, which is the third intention.	Ex hoc rursus patet tertium. Sunt enim per .13. duo anguli .a.g.e. et a.g.h. aequales duobus rectis: ergo duo anguli .a.g.h. et .c.h.g. erunt etiam aequales duobus rectis qui sunt duo intrinseci ex eadem parte sumpti: quod est tertium propositum.
Proposition 30	Propositio .30.
[English] Latin	English [Latin][¶]
If two lines will be equidistant to one, they are likewise equidistant to each other.	Si fuerint duae lineae unim aequidistantes eadem sibi invicem aequidistantes erunt.
¶ Suppose two lines, a.b and c.d, both of which are equidistant to the line e.f. I say that those two, namely a.b and c.d, are to be equidistant and this is moreover universally true whether the two lines a.b and c.d are on one surface with the line e.f or not (though here it is not perceivable unless it follows that all are on one superficies, for following that they are on separate superficies and that they are equidistant is demonstrated in the ninth of Book 11).	¶ Sint duae lineae .a.b. et .c.d. quarum utraque aequidistet lineae .e.f. dico illas duas videlicet .a.b. et .c.d. esse aequidistantes. hoc autem est universaliter verum sive duae lineae .a.b. et .c.d. sint in una super ficie cum linea .e.f. sive non: hic tamen non intelligitur nisi secundum quod omnes sunt in superficie una: secundum enim quod sunt in diversis superficiebus probatur in nona libri .11. quod sunt aequidistantes.
Then suppose they are all on one surface, and thus I will extend the line g.h cutting the lines a.b, e.f, and c.d at points k, l, and m. And because a.b is equidistant to e.f the angle b.k.l will be equal to the angle e.l.k by the first part of the preceding, since they will be coalternate. And as c.d is equidistant to e.f, the exterior angle k.l.e will equal the interior angle l.m.c by means of the second part of the preceding. Therefore, the angle b.k.l equals the angle c.m.l, which, as they are coalternate by 27, lines a.b and c.d are equidistant, which is the intent.	Sint ergo in omnes superficie una: protraham autem lineam .g.h. secantem lineas .a.b. et [sic] .c.d. in punctis .k.l.m. et quia .a.b. aequidistat .e.f. erit angulus .b.k.l. aequalis angulo .e.l.k. per primam partem pracedentis cum illi sint coalterni: atque .c.d. equidistat .e.f. erit angulus .k.l.e. extrinsecus aequalis angulo .l.m.c intrinseco per secundum partem praecedentis ergo angulus .b.k.l est aequalis angulo .c.m.l. qui cum sint coalterni erunt per .27. lineae .a.b. et .c.d. aequidistantes: quod est propositum.
Proposition 31	Propositio .31.
[English] Latin	English [Latin][¶]
To draw an equidistant line from a point outside a proposed given line.	A puncto extra lineam dato lineae proposite aequidistantem ducere.
¶ A point outside the given line is noted, wherein the line extended from both ends does not traverse through it.	¶ Punctus extra lineam datus intelligitur cum linea utrinque protrahatur per ipsum non transit.
Thus suppose the given point is a, outside the line b.c whence it behooves to extend a line equidistant to b.c. I draw out the line a.d howsoever it may contact, and at point a, which is an extremity of the line a.d, I constitute the angle e.a.d, by means of the 23rd doctrine, equal to the angle b.d.a, coalternate itself, and e.a will be equidistant to b.c by the 27th, which is the intent.	Sit ergo punctus .a. datus extra lineam .b.c. a quo oportet protrahere lineam aequidistantem .b.c. protraho lineam .a.d. qualitercunque contingat et super punctum .a. qui est extremitas lineae .a.d. constituo angulum .e.a.d. per doctrinam .23. aequalem angulo .b.d.a. sibi coalterno: eritque .e.a. aequidistans .b.c. per .27. quod est propositum.
Proposition 32	Propositio .32.
[English] Latin	English [Latin][¶]
An angle without every triangle is equal to the two interior angles opposite it. Moreover, it is necessary for all three of its angles to be equal to two right angles.	Omnis trianguli angulus exgtrinsecus duobus intrinsecis sibi oppositis est aequalis. Omnes autem tres angulos eius duobus rectis angulis equos esse necesse est.
¶ Suppose the triangle a.b.c, whose side b.c shall be extended up until point d. I say that the angle c from without is equal to the two interior angles a and b, joined opposite it, and because the three angles of the triangle a.b.c are joined together, they are equal to two right angles. From point c I will extend c.f equidistant to a.b according to the doctrine of the preceding, and the angle f.c.a will equal angle a for that they are coalternate by the first part of 29, and the angle f.c.d from without will equal the interior angle b by the second part of the very same, hence the whole of a.c.d from without is equal to a and b, the two interior angles opposite itself, which is the first intention. And because the two angles a.c.b and a.c.d are equal to two right angles by means of the 13th, the three interior angles a, b, and c, are equal to two right angles, which is the second intention.	¶ Sit triangulus .a.b.c. cuius latus .b.c. protrahatur usque ad .d. dico quod angulus .c. extrinsecus est aequalis duobus angulis .a. et .b. intrinsecis sibi oppositis simul iunctis: et quod tres anguli trianguli .a.b.c. simul iuncti sunt aequales duobus rectis. a puncto .c. protraham .c.f. aequidistantem .a.b. secundum doctrinam praecedentis: eritque angulus .f.c.a. aequalis angulo .a. quia sunt coalterni per primam partem .29. et angulus .f.c.d. extrinsecus aequalis angulo .b. intrinseco per secundum partem eiusdem: quare totus .a.c.d. extrinsecus est aequalis duobus angulis .a. et .b. intrinsecis sibi oppositis: quod est primum. et quia duo anguli .a.c.b. et .a.c.d. sunt aequales duobus rectis per .13. erunt tres anguli .a.b. et .c. intrinseci aequales duobus rectis: quod est secundum propositum.
¶ From this it is evident that all the angles of every polygonal figure taken together are equal to so many right angles, how many being double the number from that which was first. For instance:	¶ Ex hac autem patet quod omnis figurae poligoniae omnes anguli simul sumpti tot rectis angulis sunt aequales quotus est numerus quo a prima destiterit duplicatus: verbi gratia.
The triangle is the first of the polygonal figures, for if it were of two lines when a figure is to be enclosed by lines, then two right lines would be enclosing a surface, which is impossible by the last petition.	Poligoniarum figurarum est triangula prima: quia si esset duarum linearum cum figura sit clausio linearum: tunc duae lineae rectae includerent superficiem quod est impossibile per ultimam petitionem.
The quadrilateral: second; the pentagon: third. And likewise, whatever the total may be is in order by what the number of its sides or angles will be, thenceforth subtracting two.	Quadrilatera secundam: pentagona tertia. similiter autem quamlibet tota erit in ordine quotus erit numerus laterum aut angulorum eius inde dempto binario.
So, I say of the triangle, which is the first, that all of its angles are equal two right angles.	Dico ergo quod triangulae quae est prima omnes anguli sunt aequales duobus rectis.
Of the quadrilateral, which is second, theirs are equal to four right angles; and of the pentagon, which is third, theirs are equal to six right angles.	Quadrilaterae quae est secundam erunt aequales quatuor rectis: et pentagonae qui est tertia erunt aequales sex rectis.
Moreover, this is likewise evident when any such a figure is dissoluble into so many triangles (how many precisely shall be as from the first) drawing with right lines from whichever of its angles to all opposite angles. And all the angles of every triangle are equal to two right angles and all the angles of every lateral figure are equal to twice as many right angles, how many precisely shall be as from the first, which is the intent.	Hoc autem inde manifestum est quoniam cum quamlibet talis figura sit in tot triangulos resolubilis quota ipsa fuerit a prima ductis rectilineis a quovis angulorum eius ad omnes angulos oppositos: sintque omnes anguli omnis trianguli duobus rectis aequales erunt omnes lateratae figurae omnes anguli bis tot rectis aequales quota ipsa fuerit a prima: quod est propositum.
So let this be as an example:	Sit enim exempli gratia.
the pentagon is a.b.c.d.e, from whose angle a I shall draw lines to the angles c and d, opposite itself. And the whole pentagon will be resolved into three triangles, a.b.c, a.c.d, and a.d.e, where the angles of each shall equal two right angles, and the angles of the pentagon will equal six right angles (that is twice its number from which was first, or twice the number of its angles or sides thenceforth subtracting).	Pentagonus .a.b.c.d.e. a cuius angulo .a. ducam lineas ad angulos .c.d. sibi oppositos: eritque totus penthagonus resolutus in triangulos .a.b.c.a.c.d. et .a.d.e. quorum cum cuiuslibet sint anguli aequales duobus rectis erunt pentagoni anguli aequales sex rectis: quod est duplum eius numeri quo a prima distat: sive duplum numeri angulorum aut laterum eius inde dempto.
¶ And we are, what's more, able to propose the same to this effect, saying that all angles of every polygonal figure likewise accepted are equal to so many right angles; the amount is the number which is double its angles thereafter four is subtracted. So lines may be extended from whatever point marked within a figure to every angle upon the figure itself, and it shall be resolved into as many angles as are its angles, and so all the angles of those triangles likewise accepted will be equal to so many right angles, the amount being double the number of angles of the proposed figure. When in this way all the angles of the triangles into which the figure was resolved surround a center point, then they are equal to four right angles by the 13th, and the intent is manifest.	¶ Possumus quoque et sic idem proponere dicentes quod omnis figurae poligoniae omnes anguli pariter accepti sunt tot rectis angulis aequales quantus est numerus quem eius anguli duplicant inde demptis quatuor: puncto enim quolibet intra figuram signato et ab eo ad singulos angulos lineis protractis erit ipsa figura in tot angulos resoluta quanti fuerit eius anguli: ideoque omnes anguli omnius illorum triangulorum pariter accepti tot rectis anguli erunt aequales quantus est numerus quam duplicant anguli propositae figurae: cum itaque sint omnes anguli triangulorum in quos ipsa resoluta est punctum medium circumstantes quatuor rectis aequales per .13. manifestum constat propositum.
¶ Similarly, it is also evident that all the exterior angles of every polygonal figure are equal to four right angles, that indeed, they are from within and from without equal to twice as many right angles as angles they have, by means of the 13th.	¶ Similiter quoque patet quod omnis figurae poligoniae anguli omnes extrinseci quatuor rectis angulis sunt aequales: sunt enim intrinseci et extrinseci et bis tot rectis aequales quot habuerint angulos per .13.
However, the interior angles are equal to twice as many right angles as angles it has, thereafter subtracting four, thus its exterior angles are equal to four right angles, which is the intent.	Intrinseci autem sunt bis tot rectis aequales quot habuerit angulos demptis inde quatuor: ergo extrinseci sunt quatuor rectis aequales: quod est propositum.
For example: the sides of the proposed pentagon shall be extended until their angles are made exterior: a.b, indeed, extended up until f, b.c up 'til g, c.d up 'til h, d.e up 'til k, e.a up 'til l. And by the thirteenth the two angles a from within and a from without are equal to two right angles. Moreoever, by the same rational the two angles b from within and b from without, thusly and so on, whence the interior and exterior angles of a.b.c.d.e equal ten right angles. And so subtracting the interior angles that are equal to six right angles, the exterior angles, namely b.a.l, c.b.f, d.c.g, e.d.h, and a.e.k, will be equal to four right angles.	Exempli gratia: propositi penthagoni latera protrahantur ut fiant anguli extrinseci .a.b. quidem protrahatur usque ad .f.b.c. usque ad .g.c.d. usque ad .h.d.e. usque ad .k.e.a. usque ad .l. eruntque per tredecimam duo anguli .a. intrinsecus et .a. extrinsecus aequales duobus rectis: eadem autem ratione duo anguli .b. intrinsecus et .b. extrinsecus. sic et ceteri: quare .a.b.c.d.e. anguli intrinseci et extrinseci decem rectis. demptis igitur intrinsecis qui sunt aequales sex rectis erunt extrinseci videlicet .b.a.l.c.b.f.d.c.g.e.d.h. et .a.e.k. aequales quatuor rectis.
¶ Also evident is that every pentagon, each side of which cuts two from the remaining, has 5 angles equal to two right angles. Let that one such as the pentagon a.b.c.d.e be supposed, and that the side a.c shall cut the side b.e at point g and the side a.d the same side b.e at point f, and the angle a.f.g will equal the two angles b and d, since it will be from without those in the triangle f.d.b.	¶ Patet etiam quod omnis pentagoni cuius unumquodque latus duo secat ex reliquis habet .5. angulos duobus rectis aequales. sit qualis proponitur pentagonus .a.b.c.d.e. et secet latus .a.c. latus .b.e. in puncto .g. et latus .a.d. idem latus .b.e. in puncto .f. eritque angulus .a.f.g. aequalis duobus angulis .b. et .d. cum sit extrinsecus ad ipsos in triangulo .f.d.b.
And likewise, the angle f.g.a will equal the two angles c and e (since it will be exterior to those in the triangle g.c.e) but the two angles a.f.g and f.g.a along with the angle a are equal to two right angles. Therefore, the four angles b, d, c, and e are, along with the angle a, equal to two right angles, which is the intent.	Itemque angulus .f.g.a. erit aequalis duobus angulis .c. et .e. cum sit extrinsecus ad ipsos in triangulo .g.c.e. sed duo anguli .a.f.g. et .f.g.a. cum angulo .a. sunt aequales duobus rectis:ergo quatuor anguli .b.d. et .c.e. sunt cum angulo .a. aequales duobus rectis: quod est propositum.
Proposition 33	Propositio .33.
[English] Latin	English [Latin][¶]
If at the ends of two lines, equidistant and equal in size, two other lines are joined, those lines too shall be equal and equidistant.	Si summitatibus duarum linearum aequidistantium et aequalis quantitatis aliae duae lineae coniungantur ipsae quoque aequales et aequidistantes erunt.
¶ Suppose the two lines, a.b and c.d, are equal and equidistant, the ends of which shall be joined by the lines a.c and b.d. I say those are to be equal and equidistant, for I will extend the line a.d, and because the lines a.b and c.d are equidistant, the angle b.a.d will equal the angle a.d.c by the first part of 29. Therefore, the two sides, a.b and a.d, of the triangle a.b.d will be equal to the two sides, d.c and d.a, of the triangle d.c.a, and the angle d of the first will equal the angle d of the second, and so by the 4th the base b.d of the first is equal to the base a.c of the second, and the angle a.d.b of the first is equal to the angle d.a.c of the second.	¶ Sint duae lineae .a. [sic] et .b.c.d. [sic] aequales et aequidistantes quarum extremitates coniungam per lineas .a.c. et .b.d. quas dico esse aequales et aequidistantes: protraham enim lineam .a.d. et quia lineae .a.b. et .c.d. sunt aequidistantes erit angulus .b.a.d. aequalis angulo .a.d.c. per primam partem .29. ergo erunt duo latera .a.b. et .a.d. trianguli .a.b.d. aequalia duobus lateribus .d.c. et .d.a. trianguli .d.c.a. et angulus primi aequalis angulo .d. secundi: ergo per .4. basis .b.d. primi est aequalis basi .a.c. secundi et angulus .a.d.b. primi aequalis angulo .d.a.c. secundi.
And because those are coalternate, the lines b.d and a.c will be equidistant by 27, and that it was first proved they are equal, then each intention is evident.	At quia ipsi sunt coalterni erunt lineae .b.d. et .a.c. aequidistantes per .27. et quia prius probatum est ipsas esse aequales: patet propositum utrunque.
Proposition 34	Propositio .34.
[English] Latin	English [Latin][¶]
Every superficies contained by equidistant sides has equal lines and angles placed from opposite ends, and is divided through the middle by the diameter.	Omnis superficies aequidistantibus contenta lateribus lineas atque angulos ex adverso collocatos habet aequales diametro et dividente eam per medium.
¶ Suppose the superficies a.b.c.d is within equidistant sides, so that the line a.b is equidistant to c.d and a.c to b.d. I say the two lines a.b and c.d are to be just as equal as the two lines a.c and b.d, and likewise, I say the angle a is to be equal to the angle d and the angle b to the angle c. I will draw out the diameter a.d, which shall also divide the surface through its middle, for that a.b and c.d are equidistant, and the angles b.a.d and c.d.a, which are coalternate, will be equal by 29 because a.c and d.b are also equidistant, and the angles c.a.d and b.d.a, which are coalternate, are equal by means of the same.	¶ Sit superficies .a.b.c.d. aequidistantium laterum: ita quod linea .a.b. aequidistet .c.d. et .a.c.b.d. dico duas lineas .a.b. et .c.d. item duas lineas .a.c. et .b.d. esse aequales. similiter et dico angulum .a. esse aequalem angulo .d. et angulum .b. angulo .c. protraham diametrum .a.d. quae etiam dividet superficiem illam per medium cum .a.b. et .c.d. sint aequidistantes: erunt anguli .b.a.d. et .c.d.a. qui sunt coalterni aequales per .29. atquia etiam .a.c. et .d.b. sunt aequidistantes: erunt anguli .c.a.d. et .b.d.a. qui sunt coalterni aequales per eandem.
For I note the two triangles a.d.b and d.a.c, and that the two angles a and d of the triangle a.d.b are equal to the two angles d and a of the triangle d.a.c, and the side a.d that lies over each of them is common; by 26 the side a.b will be equal to the side c.d and the side a.c to the side b.d and the angle b to the angle c, and because it is evident that the entire angle a is equal to the entire angle d by the second conception: the entire proposition along with the correlation is clear.	Intelligo enim duos triangulos .a.d.b. et .d.a.c. et quia: duo anguli .a. et .d. trianguli .a.d.b. sunt aequales duobus angulis .d. et .a. trianguli .d.a.c. et latus .a.d. super quod iacent illi anguli in utroque triangulo est commune: erit per .26. latus .a.b. aequale lateri .c.d. et latus .a.c. lateri .b.d. et angulus .b. angulo .c. et quia angulum .a. totalem patet esse aequalem angulo .d. totali per secundam conceptionem: totum propositum cum correlatio liquet.
Proposition 35	Propositio .35.
[English] Latin	English [Latin][¶]
All superficies of equidistant sides constituted upon one base and in the selfsame alternate lines are proved to be equal.	Omnes superficies aequidistantium laterum super unam basim atque in eisdem alternis lineis constitutae aequales esse probantur:
¶ Suppose the two lines a.b and c.d are equidistant, between which suppose a.c.f.e is a superficies of equidistant sides over the base c.e, and suppose over the same base and between the same lines is another superficies, g.c.h.e, likewise of equidistant sides. I say the two predicated superficies are to be equal, which is proved thus: for either the line c.g will cut the line a.b at any point on the line .a.f, or at point f, or at any point upon the line b.f. And so first it shall cut at any point upon the line a.f, as appears within the first figure. And because both of the lines a.f and g.h are to be equal to the line c.e, through the preceding one of them will equal the other. Then by removing the common line f.g, a.g will remain equal to f.h because, again by the preceding, a.c is equal to f.e, and the angle h.f.e to the angle g.a.c by the second part of 29, namely, one from without will be to one within, and by means of the 4th, the triangle a.c.g will equal the triangle f.e.h. Therefore, the irregular quadrilateral figure, which is g.c.f.e, will be added to both and the superficies a.c.f.e will equal the surface g.c.h.e, which is the intent.	¶ Sint duae lineae .a.b. et .c.d. aequidistantes inter quas fiat .a.c.f.e. superficies aequidistantium laterum super basim .c.e. et super eandem basim et inter easdem lineas fiat alias superficies .g.c.h.e. similiter aequidistantium laterum: dico duas praedictas superficies esse aequales: quod sic probatur aut enim linea .c.g. secabit lineam .a.b. in aliquo puncto lineae .a.f. aut in puncto .f. aut in aliquo puncto lineae .b.f. secet ergo primo in aliquo puncto lineae .a.f. ut in prima figurationae apparet. et quia utraque duarum linearum .a.f. et .g.h. esse aequalis lineae c.e. per praecedentem una earum erit aequalis alteri dempta ergo linea .f.g. commmuni remanebit .a.g. aequalis .f.h. quia per praecedentem iterum est .a.c. aequalis .f.e. et angulus .h.f.e. angulo .g.a.c. per secundam partem .29. videlicet extrinsecus intrinseco erit per .4. triangulus .a.c.g. aequalis triangulo .f.e.h. ergo irregulari figura quadrilatera quae est .g.c.f.e. addita utrique erit superficies .a.c.f.e. aequalis superficiei .g.c.h.e. quod est propositum.
And so in this way the line c.g shall cut the line a.b at point f, as is apparent in the second figure, and by a similar argument to the former, the two triangles, a.c.f and f.e.h, will be equal, wherefore by the addition of the triangle f.c.e to both, the proposition is evident.	Secet ergo modo linea .c.g. lineam .a.b. in puncto .f. ut in secunda figuratione apparet. eruntque simili argumentatione priori duo trianguli .a.c.f. et .f.e.h. aequales: quare utrobique addito triangulo .f.c.e. patet propositum.
In a third way, the line c.g shall cut the line a.b between the two points f and b, as is apparent in the third figure, then it will cut the line f.e thus at point k and because it is from a similar reasoning to the prior, the line a.f is equal to the line g.h, and being made from the common line g.f, a.g will be equal to f.h, and the triangle a.g.c will be equal to the triangle f.e.h. Therefore, adding to both the triangle c.k.e and subtracting from both the triangle f.k.g, the surface a.c.f.e will equal the superficies g.c.h.e, which is the intent.	Secet tertio modo linea .c.g. lineam .a.b. inter duo puncta .f.b. ut in tertia figuratione apparet: secabitque lineam .f.e. sic ut in puncto .k. et quia simili argumentatione priori linea .a.f. est aequalis lineae .g.h. facta communi linea .g.f. erit .a.g. aequalis .f.h. et triangulus .a.g.c. aequalis triangulo .f.e.h. additio ergo utrique triangulo .c.k.e. et detracto ab utroque triangulo .f.k.g. erit superficies .a.c.f.e. aequalis superficiei .g.c.h.e. quod est propositum.
Proposition 36	Propositio .36.
[English] Latin	English [Latin][¶]
It is necessary that all parallelograms constituted upon equal bases and in the selfsame lines are to be equal.	Omnia paralellograma in basibus aequalibus atque in eisdem lineis constituta aequalia esse necesse est.
¶ It is said the parallelogram is a superficies of equidistant sides.	¶ Paralellogramum dicitur superficies aequidistantium laterum.
Suppose the two superficies, a.b.c.d and e.f.g.h, are constituted of equidistant sides between two equidistant lines, which are a.f and c.h, and they are on equal bases, which are c.d and g.h. I say they are to be equal, for I will draw out the two lines c.e and d.f and by 33, the surface c.d.e.f will be of equidistant sides and on account of that, e.f is equal and equidistant to c.d since both of them are equal to g.h. And so by the preceding, that both of the superficies a.b.c.d and e.f.g.h are equal to the surface c.d.e.f, they will be mutually equal to one another, which is the intent.	Sint duae superficies .a.b.c.d. et .e.f.g.h. aequidistantium laterum constitutae inter duas lineas aequidistantes quae sunt .a.f. et .c.h. et super aequales bases quae sunt .c.d. et .g.h. dico eas esse aequales. nam protraham duas lineas .c.e. et .d.f. eritque per .33. superficies .c.d.e.f. aequidistantium laterum propter hoc quod .e.f. est aequalis et aequidistans .c.d. nam utraque earum est aequalis .g.h. quia ergo per praemissam utraque duarum superficierum .a.b.c.d. et .e.f.g.h. est aequaqlis superficiei .c.d.e.f. ipsi erunt sibi invicem aequales: quod est propositum.
Proposition 37	Propositio .37.
[English] Latin	English [Latin][¶]
All triangles are wholly equal to one another that are constructed upon the same base and between two equidistant lines.	Aequales sunt sibi cuncti trianguli qui super eandem basim atque inter duas lineas aequidistantes sunt constituti.
¶ Suppose the two triangles a.b.c and d.b.c are constituted over the base b.c, between the two lines a.e and b.f, of which are equidistant. I say they are to be equal, for I will draw out c.g equidistant to a.b and c.g equidistant to d.b by 31, and the two superficies a.b.c.g and d.b.c.h will be equal by 35. And because the stated triangles are half of those, through a corollary of the 34th, they will be equal through the common knowledge, which is that the wholes of them are equal as well their halves, and thus the proposition is evident.	¶ Sint duo trianguli .a.b.c. et .d.b.c. constitutae super basim .b.c. inter duas lineas .a.e. et .b.f. quae sunt aequidistantes dico eas esse aequales protraham enim .c.g. aequidistantem .a.b. et .c.h. aequidistantem .d.b. per .31. eruntque duae superficies .a.b.c.g. et .d.b.c.h. aequales per .35. et quia dicti trianguli sunt earum dimidia per correlarium .34. ipsae erunt aequales per communem scientiam: quae est quorum tota sunt aequalia et dimidia: sicque patet propositum.
Proposition 38	Propositio .38.
[English] Latin	English [Latin][¶]
If two triangles should fall upon equal bases as well between two equidistant lines, it is necessary they are to be equal.	Si duo trianguli super bases aequales atque inter duas lineas aequidisatantes ceciderint aequales eos esse necesse est.
¶ Suppose the two triangles a.b.c and d.e.f are constituted over the equal bases b.c and c.f, and between the equidistant lines a.g and b.h. I say they are to be equal, for I will draw c.k equidistant to a.b and f.l equidistant to e.d, and the two superficies a.b.c.k and d.e.f.l will be equal by 36. And because the stated triangles are half of them through a corollary of the 34th, they will be equal through aforesaid common knowledge.	¶ Sint duo trianguli .a.b.c. et .d.e.f. constituti super bases .b.c. et .e.f. aequales et inter lineas .a.g. et .b.h. aequidistantes: dico eos esse aequales. protraham enim .c.k. aequidistantem .a.b. et .f.l. aequidistantem .e.d. eruntque duae superficies .a.b.c.k. et .d.e.f.l. aequales per .36. et quia dicti trianguli sunt earum dimidia per correlarium .34. ipsi erunt aequales per antedictam communem scientiam.
Proposition 39	Propositio .39.
[English] Latin	English [Latin][¶]
All two equal triangles, if they should fall upon the same base and from the same side, will be between two equidistant lines.	Omnes duo trianguli aequales si in eandem basim et ex eadem parte ceciderint: inter duas lineas aequidistantes erunt.
¶ Suppose the two triangles a.b.c and d.b.c are constituted over the base b.c, both from the same side, and that they are equal. I say then they are to be between equidistant lines, and this is a converse of 37. From point a I will extend a line equidistant to the line b.c, which, if it shall traverse by way of point d, the proposition is clear.	¶ Sint duo trianguli .a.b.c. et .d.e.f. [sic] constituti super basim .b.c. ex una eademque parte: sintque aequales: dico eas esse inter lineas aequidistantes: et haec est conversa .37. a puncto .a. protraham lineam aequidistantem lineae .b.c. quae si per transierit per punctum .d. liquet propositum.
If, however, it should pass over or below, then let it first pass over and be as a.e. And I will draw b.d up until it shall cut the line a.e at point e, and I will draw out the line e.c, and because the triangle e.b.c is equal to the triangle a.b.c by 37 and the triangle d.b.c is positioned equal to triangle a.b.c, the triangle d.b.c will be equal to the triangle e.b.c, a part of the whole, which is impossible.	Si autem pertransierit supra aut infra: transeat primo supra et sit .a.e. producamque .b.d. usquequo secet lineam .a.e. in puncto .e. et protraham lineam .e.c. et quia triangulus .e.b.c. est aequalis triangulo .a.b.c. per .37. et triangulus .d.b.c. positus est aequalis triangulo .a.b.c. erit triangulus .d.b.c. aequalis triangulo .e.b.c. pars toti quod est impossibile.
Thus a line which is drawn equidistant to b.c from point a will not pass over d, then let it pass below and suppose it is as a.f, cutting the line d.b at point f. And I will draw out the line f.c, and because by 37 the triangle f.b.c is equal to the triangle a.b.c, it will also be equal to the triangle d.b.c, a part of the whole, which is impossible.	Non igitur protransibit linea quae a puncto .a. ducitur aequidistanter .b.c. supra .d. transeat ergo infra et sit .a.f. secans lineam d.b. in puncto .f. protraham ergo lineam .f.c. et quia per .37. triangulus .f.b.c. est aequalis triangulo .a.b.c. ipse etiam erit aequalis tirangulo .d.b.c. pars toti quod est impossibile.
Therefore, that a line does not pass equidistant to b.c from point a if not through point d, the proposition is evident.	Quia ergo linea a puncto .a. aequidistanter .b.c. non transit nisi per punctum .d. patet propositum.
¶ From this, however, and going forward, note that if any right line shall cut or will cut any two sides of a triangle by equals, the third side will be equidistant, which is proved as follows:	¶ Ex haec autem et praemissa nota quod si aliqua linea recta duo alicuius trianguli latera per aequa secet vel secuerit ipsa erit tertio aequidistans quod sic probatur.
Suppose the triangle a.b.c, whose two sides, which are a.b and b.c, shall cut the line d.e by equals, that is, a.b at point d and b.c at point e. I say that the line d.e. is equidistant to a.c. For I will draw in the quadrilateral a.c.e.d the diameters a.e and d.c, and by 38 the triangle a.e.d will equal the triangle d.e.b, and because the line a.d is equally positioned to the line d.b, likewise and by the same, the triangle c.e.d will equal that same triangle d.e.b. On account of that, the line c.e is positioned equally to e.b because the triangle a.e.d is equal to the triangle c.e.d, and so since they are constructed upon the same base, namely, the line e.d, and they are from the same side, then they are by this, the 39th, between equidistant lines, and thus the line d.e is equidistant to the line a.c. A proposition that will, in fact, be useful to you at the fifth of the fourth.	Sit triangulus .a.b.c. cuius duo latera quae sunt .a.b. et .b.c. secet lineam .d.e. per aequalia .a.b. quod in puncto .d. et .b.c. in puncto .e. dico quod linea .d.e. est aequidistans .a.c. protraham enim in quadrilatero .a.c.e.d. diametros .a.e. et .d.c. eritque per .38. triangulus .a.e.d. aequalis triangulo .d.e.b. propter id quod linea .a.d. posita est aequalis lineae .d.b. itemque per eandem triangulus .c.e.d. erit aequalis idem triangulo .d.e.b. propter id quod linea .c.e. posita est aequalis lineae .e.b. quia triangulus .a.e.d. est aequalis triangulo .c.e.d. quia ergo ipsi sunt constituti super eandem basim videlicet lineam .e.d. et ex eadem parte ipsi erunt per hanc .39. inter lineas aequidistantes ergo linea .d.e. est aequidistans lineae .a.c. quod quidem propositum ad quintam quarti tibi valebit.
Proposition 40	Propositio .40.
[English] Latin	English [Latin][¶]
If two equal triangles shall be constituted upon equal bases and from the same side of one line, it is necessary they should be contained between two equidistant lines.	Si duo trianguli aequales super aequales bases unius eiusdemque lineae ex eadem parte fuerint constituti eos inter duas lineas aequidistantes necesse est contineri.
¶ Suppose the two triangles a.b.c and d.e.f are constituted over the two bases, which are b.c and e.f, and out from the same side. I say they are to be between two equidistant lines, and this is converse the 38th, and it is proved by that, just as the preceding is by 37. From point a, suppose a line is drawn equidistant to the line b.f, which, if it shall traverse by way of point d, the proposition is evident.	¶ Sint duo trianguli .a.b.c.d.e.f. aequales constituti super duas bases quae sunt .b.c. et .e.f. et ex eadem partem dico eos esse inter duas lineas aequidistantes et haec est conversa .38. et probatur per ipsam sicut praecedens per .37. a puncto .a. ducatur linea aequidistans lineae .b.f. quae si transierit per punctum .d. patet propositum.
However if on the contrary it should pass over as a.g, and .e.d is extended up until that, which is .e.g, and the line d.f is drawn, then by 38 the triangle a.b.c will equal the triangle g.e.f, and so the triangle d.ef will equal the triangle g.e.f, a part of the whole, which is impossible, and so it will not pass over. Suppose it should go below, and it shall cut the line d.e at point h and the line f.h is drawn. And by 38 the triangle b.e.f equals the triangle a.b.c, thus the triangle d.e.f, a part of the whole, which is impossible. Therefore, that it will not pass if not through point d, the proposition is evident.	Sin autem pertransierit supra ut .a.g. et producatur .e.d. usque ad ipsum quae sit .e.g. et ducatur linea .g.f. eritque per .38. triangulus .a.b.c. aequalis triangulo .g.e.f. quare et triangulus .d.e.f. erit aequalis triangulo .g.e.f. pars toti quod est impossibile. Non ergo transibit supra: transeat ergo infra. et secet lineam .d.e. in puncto .h. et ducatur linea .f.h. eritque per .38. triangulus .b.e.f. aequalis triangulo .a.b.c. quare et triangulo .d.e.f. pars toti quod est impossibile. quia ergo non transibit nisi per punctum .d. patet propositum.
Proposition 41	Propositio .41.
[English] Latin	English [Latin][¶]
If a parallelogram and triangle should be constituted on the same base and within the same alternate lines, the parallelogram shall aptly be double the triangle.	Si paralellogramum triangulusque in eadem basi atque in eisdem alternis lineis fuerint constituta paralellogramum triangulo duplum esse conveniet.
¶ Suppose the parallelogram a.b.c.d and the triangle e.b.d are over the base b.d and between the lines a.c and b.d, which are equidistant. I say the parallelogram is to be double the triangle. I will draw out within the parallelogram the diameter a.d and the triangle a.b.d will be half the parallelogram through a corollary of 34, and because the triangle e.b.d is equal to the triangle a.b.d, by 37 it is evident the triangle e.b.d is to be half the parallelogram a.b.c.d, which is the intent.	¶ Sit paralellogramum .a.b.c.d. et triangulus .e.b.d. super basim .b.d. et inter lineas .a.c. et .b.d. qui sint aequidistantes: dico paralellogramum duplum esse triangulo: protraham in paralellogramo diametrum .a.d. eritque triangulus .a.b.d. dimidium paralellogrami per corellarium .34. et quia triangulus .e.b.d. est aequalis triangulo .a.b.d. per .37. patet triangulum .e.b.d. esse dimidium paralellogrami .a.b.c.d. quod est propositum.
Similarly, it is also be proved that if a parallelogram and triangle are constituted on equal bases and between equidistant lines, the parallelogram will be double the triangle. That Euclid did not posit such is because it is somewhat evident by the preceding corollary and by 38, by dividing the parallelogram through the diameter into two triangles; or by a triangle constructed over the base of a parallelogram, between the same equidistant lines, to which the parallelogram will be double by this preceding, and that [triangle] will equal the other triangle by 38.	Similiter quoque potest probari quod si paralellogramum triangulusque in aequalibus basibus atque inter lineas aequidistantes fuerint constituta paralellogramum duplum erit triangulo: quod ideo non posuit euclides quia leviter patet ex haec praecedente corollarium et .38. diviso paralellogramo per diametrum in duos triangulos. vel super basim paralellogrami inter easdem lineas aequidistantes triangulo constituto ad quae duplum erit paralellogramum per hanc praecedentem et ipse aequalis alteri triangulo per .38.
Proposition 42	Propositio .42.
[English] Latin	English [Latin][¶]
To designate a surface of equidistant sides whose angle shall equal an assigned angle, and the superficies itself shall equal an assigned triangle.	Aequidistantium laterum superficiem designare cuius angulus sit angulo assignato aequalis. ipsa vero superficies triangulo assignator aequalis.
¶ Suppose the assigned angle is a and the assigned triangle is b.d.c. I mean to describe a surface of equidistant sides equal to the triangle b.c.d, of whose two angles of opposite positions to one another are each equal to a. I divide the base c.d by half at point e and I extend the line b.e, and from point b draw b.f, equidistant to c.d, and through 38, the triangle b.e.d will equal the triangle b.e.c, whereby the triangle b.e.d is half the total of the triangle b.c.d. Thus over point e of the line d.c, I constitute the angle d.e.g, equal to angle a, and I finish the parallelogram g.e.d.f, which too, that by the preceding is double the triangle b.e.d and furthermore will equal the triangle b.c.d through this common knowledge, that the halves of those which are equal are themselves also equal. And indeed, the triangle b.e.d is half of each, by which we have described the parallelogram g.e.d.f equal to the triangle b.c.d, each of whose two angles, g.e.d and d.f.g of opposite positions to the other, are equal to angle a, which was the intent.	¶ Sit assignatus angulus .a. et assignaturs triangulus .b.c.d. volo describere superficiem aequidistantium laterum aequalem triangulo .b.c.d. cuius uterque duorum angulorum contra se positorum sit aequalis .a. divido basim .c.d. per dimidium in puncto .e. et protraho lineam .b.e. et a puncto .b. duco .b.f. aequidistantem .c.d. eritque per .38. triangulus .b.e.d. aequalis triangulo .b.e.c. quare triangulus .b.e.d. est dimidium totalis trianguli .b.c.d. igitur super punctum .e. lineae .d.c. constituo angulum .d.e.g. aequalem angulo .a. et perficio paralellogramum .g.e.d.f. quod etiam quia per praecedentem est duplum ad triangulum .b.e.d. erit etiam aequale triangulo .b.c.d. per haec communem scientiam: quorum dimidia sunt aequalia ipsa quoque sunt aequalia. est enim triangulus .b.e.d. utriusque dimidium quare descripsimus paralellogramum .g.e.d.f. aequale triangulo .b.c.d. cuius uterque duorum angulorum .g.e.d. et .d.e.f. contra se positorum est aequalis angulo .a. quod fuit propositum.
Proposition 43	Propositio .43.
[English] Latin	English [Latin][¶]
It is necessary of the space of every parallelogram for those supplements of the parallelograms that are about the diameter to be mutually equal.	Omnis paralellogrami spacii eorum quae circa diametrum sunt paralellogramorum supplementa aequa sibi invicem esse necesse est.
¶ Suppose the parallelogram is a.b.c.d, in which I will draw out the diameter b.c and I will draw out e.f equidistant to both of the two sides, a.b and c.d, which shall cut the diameter at point k, from which I shall draw k.g equidistant to each of the two sides, a.c and b.d, and that I will lengthen until it shall cut both the side a.b and c.d, and the total shall be g.k.h. For the whole parallelogram a.b.c.d will be divided into four parallelograms, two of which, namely, e.c.k.h and g.k.b.f, are said to be located about c.b because the diameter passes through the middle of them (and thus they are about the diameter) and the remaining two, namely a.e.g.k and k.h.f.d, are said to be supplements, and the two supplements here are said to be equal, for the two triangles a.b.c and c.d.b are equal by a corollary of 34. Similarly, the two triangles, g.k.b and f.k.b, are as well equal by the selfsame 34th corollary.	¶ Sit paralellogramum .a.b.c.d. in quo protraham diametrum .b.c. et protraham .e.f. aequidistantem uterque duorum laterum .a.b. et .c.d. quae secet diametrum in puncto .k. a quo ducam .k.g. aequidistant est utrique duorum laterum .a.c. et .b.d. et producam eam quousque secet utrumque latus .a.b. et .c.d. sitque tota .g.k.h. erit quia totum paralellogramum .a.b.c.d. divisum in quatuor paralellograma quorum duo scilicet .e.c.k.h. et .g.k.b.f. dicuntur consistere circa .c.b. quod diametrum transit per medium eorum et ideo sunt circa diametrum: reliqua duo scilicet .a.e.g.k. et .k.h.f.d. dicuntur supplementa haec duo supplementa dicuntur esse aequalia. sunt enim duo trianguli .a.b.c. et .c.d.b. aequales per correlarium .34. similiter quoque duo trianguli .g.k.b. et .f.k.b. sunt aequales per idem correlarium .34.
And the two triangles c.e.k and k.h.c are likewise equal by the same corollary, thus by removing the two triangles b.g.k and k.e.c from the whole of the triangle a.b.c, and the two remaining triangles b.f.k and k.c.h from whole of the remaining triangle c.d.b, through common knowledge those will remain, which are the two so-stated supplements, are equal, which is the intent.	At duo trianguli .c.e.k. et .k.h.c. similiter aequales per idem correlarium demtpis igitur duobus triangulis .b.g.k. et .k.e.c. de totali triangulo .a.b.c. ac duobus triangulis reliquis .b.f.k. et k.c.h. de totali triangulo reliquo .c.d.b. erunt per communem scientiam residua: quae sunt duo dicta supplementa aequalia: quod est propositum.
Proposition 44	Propositio .44.
[English] Latin	English [Latin][¶]
Propose a right line over which to designate a surface of equidistant sides, whose angle shall equal an assigned angle but whose very superficies shall equal an assigned triangle.	Proposita linea recta super eam superficiem aequidistantium laterum cuius angulus sit angulo assignato aequalis ipsa vero superficies triangulo assignato aequalis designare.
¶ To designate a surface of equidistant sides upon any line, is to make that line one side of its surface.	¶ Designare superficiem aequidistantium laterum super linea aliquam est lineam ipsam facere latus unum ipsius superficiei.
Therefore, suppose a.b is the given line, and c the given angle, and d.e.f the given triangle. I wish to designate one superficies of equidistant sides over the line a.b, so that the line a.b shall be one of its sides, whereof each of the two angles in positions opposite itself shall equal the angle c, and its whole surface shall equal the triangle d.e.f. This, however, differs from 42 because here, one given side of the surface is described, namely, the line a.b, whereas there, none are. So, when I mean to do this, I join the line a.b to the line a.g according to rectitude, whereby I place it equal to the line e.f, the base of the given triangle, over which I constitute one triangle equal and equilateral to it, which I do in this way:	Sit ergo data linea .a.b. et datus angulus .c. et datus triangulus .d.e.f. super lineam .a.b. volo designare superficiem unam aequidistantium laterum ita quod linea .a.b. sit unum ex lateriubus eius cuius uterque duorum angulorum contra se positorum sit aequalis angulo .c. et ipsa totalis superficies sit aequalis triangulo .d.e.f. differt autem haec a .42. quia hic datus latus unius superficiei describendae scilicet linea .a.b. ibi autem nullum. cum ergo voluero facere adiungo lineam .a.g. lineae .a.b. secundum rectitudinem: quam pono aequalem lineae .e.f. basi trianguli dati super quam constituo triangulum unum ei aequales et aequilaterum. quod hoc modo facio.
I constitute the angle a.g.k equal to angle e, and angle g.a.k equal to angle f by 23, and because g.a will be positioned equal to e.f, by 26 the triangle g.a.k is equilateral and equal to the triangle e.f.d. Thus I will divide g.a by equals at point h, and I will extend k.h, and I will extend the line m.k.n from point k, equidistant to the line g.b. And by 38, the triangle a.h.k will equal the triangle g.h.k. Then over point a of the line g.a, I will make angle g.a.l, by means of 23, equal to the given angle c, and over the base a.h and between the equidistant lines g.b and m.n, I will complete a surface of equidistant sides, m.l.h.a, which by 41, will be double the triangle h.k.a, whereby it is equal to the whole triangle k.g.a and so to the proposed triangle .d.e.f. Therefore I will extend b.n equidistant to a.l and draw out the diameter n.a, which I will extend until it shall coincide with m.h at point o, and I will complete the surface of equidistant sides m.o.n.q, and extend l.a up until p. And by the preceding the supplement a.b.p.q will equal the supplement m.l.h.a, and so too the triangle d.e.f. And because by the 15th the angle l.a.h is equal to the angle b.a.p and so the angle b.a.p is equal to the angle c, it is evident the surface of equidistant sides described over the given line is to be equal to the given triangle d.e.f, each of whose two angles in positions opposite it, which are a and g, are equal to the given angle c, which was the intent.	Constituo angulum .a.g.k. aequalem angulo .e. et angulum .g.a.k aequalem angulo .f. per .23. et quia .g.a. posita fuerat aequalis .e.f. erit per .26. triangulus .g.a.k. aequalis et aequilaterus triangulo .e.f.d. dividam ergo .g.a. per aequalia in puncto .h. et protraham .k.h. et producam a puncto .k. lineam .m.k.n. aequidistantem lineae .g.b. eritque per .38. triangulus .a.h.k. aequalis triangulo .g.h.k. tunc super punctum .a. lineae .g.a. faciam anglum .g.a.l. per .23. aequalem angulo .c. dato: et conplebo super basim .a.h. et inter lineas .g.b. et .m.n. aequidistantes superficiem aequidistantius laterum .m.l.h.a. quae per .41. dupla erit ad triangulum .k.h.a. quare aequalis totali triangulo .k.g.a. quare et triangulo .d.e.f. proposito: protraham ergo .b.n. aequidistantem .a.l. et producam diametrum .n.a. quam protraham quousque concurrat cum .m.h. in puncto .o. et conplebo superficiem aequidistantium laterum .m.o.n.q. et protraham .l.a. usque ad .p. eritque per praecedentem supplementum .a.b.p.q. aequale supplemento .m.l.h.a. quare et triangulo .d.e.f. et quia per .15. angulus .l.a.h. est aequalis angulo .b.a.p. et ideo angulus .b.a.p. est aequalis angulo .c. patet super datam lineam .a.b. descriptam esse superficiem aequidistantium laterum .a.b.p.q. aequalem dato triangulo .d.e.f. cuius uterque duorum angulorum contra se positorum qui sunt .a. et .q. est aequqalis dato angulo .c. quod fuit propositum.
Proposition 45	Propositio .45.
[English] Latin	English [Latin][¶]
To describe a quadrate from a given line.	Ex data linea quadratum describere.
¶ Suppose the given line from which I mean to describe a quadrate is a.b. From the points a and b of the line a.b, I draw by the 11th the lines a.c and b.d perpendicular to the line a.b, which are equidistant through the last part of 28. And I place both of them equal to the same a.b by the following, and extend the line c.d, and that will be equal and equidistant to the line a.b through the 33rd. and because each of the two angles a and b are right, each of the two, c and d, are right by the last part of 29, and so a.b.c.d is a quadrate by definition, which is the intent.	¶ Sit data linea .a.b. ex qua volo quadratum describere: a punctis .a. et .b. lineae .a.b. educo per .11. lineas .a.c. et .b.d. perpendiculares ad lineam .a.b. quae erunt aequidistantes per ultimam partem .28. et pono utramque earum eidem .a.b. per secundam aequalem et protraho lineam .c.d. eritque ipsa aequalis et aequidistans lineae .a.b. per .33. et quia uterque duorum angulorum .a. et .b. est rectus. erit uterque duorum .c. et .d. rectus per ultimam partem .29. ergo per diffinitionem .a.b.c.d. est quadratum quod est propositum.
Otherwise suppose the same a.c is perpendicular to the line a.b by means of the 11th, and let it be equal to it as the first, and from point c, by 31, c.d is drawn equidistant to a.b and placed equal to it, and the line d.b is drawn, which, by 33, will be equidistant and will equal a.c, and all angles right by the last part of 29, where by definition we have the intent.	Idem aliter sit .a.c. perpendicularis super lineam .a.b. per .11. et sit ei aequalis ut prius et a puncto .c. per .31. ducatur c.d. aequidistans .a.b. et ponatur aequalis ei et ducatur linea .d.b. quae per .33. erit aequalis et aequidistans .a.c. et omnes anguli recti per ultimam partem .29. quare per diffinitionem habemus propositum.
Proposition 46	Propositio .46.
[English] Latin	English [Latin][¶]
In every right-angled triangle, the quadrate that is drawn from the side opposite the right angle is itself equal to the two quadrates which are drawn out from the two remaining sides.	In omni triangulo rectangulo quadratum quod a latere recto angulo opposito in semetipso ducto describitur aequum est duobus quadratis quae ex duobus reliquis lateribus conscribuntur.
¶ Suppose the triangle is a.b.c, of whose angle a is right. I say that the quadrate of the side b.c is equal to quadrate a.b and quadrate a.c, taken together.	¶ Sit triangulus .a.b.c. cuius angulus .a. sit rectus dico quod quadratum lateris .b.c. aequum est quadrato .a.b. et quadrato .a.c. simul sumptis.
And so I will square these three sides according to the preceding doctrine, and suppose the quadrate b.c is the superficies b.c.d.e, and quadrate b.a, the superficies b.f.g.a, and quadrate a.c, the superficies a.c.h.k. From the right angle a, I will draw three lines to the base d.e, the base of the greatest quadrate, namely: a.l, equidistant to each side, b.d and c.e, which shall cut b.c at point m of the hypotenuse, and a.d, and a.e. Likewise, from the two remaining angles of the triangle, which are b and c, I will draw two lines to the two angles of the two smaller quadrates that intersect one another within the triangle itself, which are b.k and c.f. And because both of the two angles, b.a.c and b.a.g, are right, by the 14th g.c will be one line. By the same rational, b.h will be one line because both of the two angles, c.a.b and c.a.h, are right. Thus, that the parallelogram b.f.g.a and the triangle b.f.c are constructed over the base b.f and between two equidistant lines, which are c.g and b.f, by 41, the parallelogram b.f.g.a will be double the triangle b.f.c. To wit, the triangle b.f.c is equal to the triangle b.a.d by the 4th, because the sides f.b and b.c of the first are equal to a.b and b.d, the sides of the last, and the angle b of the first is equal to angle b of the last, for each is composed of a right angle and common angle, a.b.c. So the parallelogram b.f.g.a is double the triangle a.b.d, but the parallelogram b.d.l.m is double that same triangle by 41, because they are constituted over the same base, namely, b.d, and between equidistant lines, which are b.d and a.l. Therefore, through common knowledge, the quadrate a.b.f.g and the parallelogram b.d.l.m are equal, because their halves, namely, the aforesaid triangles, are equal. In the same way and through the same propositions mediating the triangles k.b.c and a.e.c, we may demonstrate the quadrate a.c.h.k is to be equal to the parallelogram c.e.l.m, whereby the proposition is evident.	Quadrabo ergo haec tria latera secundum doctrinam praecedentis: sitque quadratum .b.c. superficies .b.c.d.e. et quadratum .b.a. superficies .b.f.g.a. et quadratum .a.c. superficies .a.c.h.k. ab angulo .a. recto ducam ad basim .d.e. basim maximi quadrati tres lineas scilicet .a.l. aequidistantem utrique lateri .b.d. et .c.e. quae secet b.c. in puncto .m. et ypothemisas .a.d. et .a.e. itemque a duobus reliquis angulis trianguli qui sunt .b. et .c. ducam ad duos angulos duorum quadratorum miorum duas lineas se intersecantes intra ipsum triangulum quae sunt .b.k. et .c.f. et quia uterque duorum angulorum .b.a.c. et .b.a.g. est rectus per .14. erit .g.c. linea una: eadem ratione erit .b.h. linea una. quia uterque duorum angulorum .c.a.b. et .c.a.h. est rectus: quia ergo super basim .b.f. et inter duas lineas aequidistantes quae sunt .c.g. et .b.f. constituta sunt paralellogramum .b.f.g.a. et triangulus .b.f.c. erit per .41: paralellogramum .b.f.g.a. duplum triangulo .b.f.c. scilicet triangulus .b.f.c. est aequalis triangulo .b.a.d. per .4. quia .f.b. et .b.c. latera primi sunt aequalia .a.b. et .b.d. lateribus postremi. et angulus .b. primi est aequalis angulo .b. postremi. eo quod uterque constat ex angulo recto et angulo .a.b.c. communi. ergo paralellogramum .b.f.g.a. est duplum ad triangulus .a.b.d. sed paralellogramum .b.d.l.m. est duplum ad eundem triangulum per .41. quia constituti sunt super eandem basim scilicet .b.d. et inter lineas aequidistnates quae sunt .b.d. et .a.l. ergo per communem scientiam quadratum .a.b.f.g. et paralellogramum .b.d.l.m sunt aequalia. quia eorum dimidia videlicet praedicti trianguli sunt aequalia. Eodem modo et per easdem propositiones mediantibus triangulis .k.b.c. et .a.e.c. probabimus quadratum .a.c.h.k. esse aequale paralellogramo .c.e.l.m. quare patet propositum.
Proposition 47	Propositio .47.
[English] Latin	English [Latin][¶]
If that which is produced by one side of a triangle drawn according to itself shall be equal to the two quadrates that are described from the two remaining sides, the angle to which that side is opposite, is right.	Si quod ab uno trianguli latere in seipsum ducto producitur: aequum fuerit duobus quadratis quae a duobus reliquis lateribus describuntur. rectus est angulus cui latus illud opponitur.
¶ To draw a line is in itself to describe its very quadrate.	¶ Lineam in se ipsam ducere est eius quadratum describere.
¶ Suppose the triangle is a.b.c, and let the quadrate of the side a.c equal the quadrates of the two sides a.b and b.c, joined together. I say the angle b, which is opposite the side a.c, is to be right. And this is a converse of the former.	¶ Sit triangulus .a.b.c. sitque quadratum lateris .a.c. aequale quadratis duorum laterum .a.b. et .b.c. simul iunctis. dico angulum .b. cui latus .a.c. opponitur esse rectum: et haec est conversa prioris.
¶ From point b I draw out the line b.d, by the 11th, perpendicular to the line b.c, which I place equal to a.b. And I draw the line d.c, and by the preceding, the quadrate d.c will equal the two quadrates of the two lines, d.b and b.c. And because its position is equal to b.a, through common knowledge, namely that equal quadrates are to be of equal lines, the quadrates of the two lines a.b and b.d are equal, wherefore the quadrate d.c will equal the quadrate of a.c. Thus, by yet other common knowledge that is a converse of the former, namely, lines whose quadrates are equal are to be equal, then d.c will be equal to a.c whereby through the 8th, the angle b of the triangle a.b.c is right, which is the intent.	¶ A puncto .b. extraho lineam .b.d. per .11. perpendicularem super lineam .b.c. quam pono aequalem .a.b. et produco lineam .d.c. eritque per praecedentem quadratum .d.c. aequale duobus quadratis duarum linearum .d.b. et .b.c. et quia .b.d. posita est aequalis .b.a. erunt per communem scientiam quae est linearum aequalium aequalia esse quadrata: quadrata duarum linearum .a.b. et .b.d. aequalia: quapropter erit quadratum .d.c. aequale quadrato .a.c. ergo per aliam communem scientiam quae est conversa prioris scilicet lineas quarum quadrata sunt aequalia esse aequales: erit .d.c. aequalis a.c. quare per .8. angulus .b. trianguli .a.b.c. est rectus quod est propositum.
Proposition 48	Propositio .48.
[English] Latin	English [Latin][¶]
For whichsoe'er quadrates proposed, to draw out upon one of them a gnomon equal to the remaining.	Propositis quibuscunque quadratis alteri illorum gnomonem reliquo aequalem describere.
¶ Two quadrates are thus proposed, namely, a.b and c.d, and suppose the intent is to produce a gnomon about a.b equal to the quadrate c.d, and so one side of the quadrate a.b shall be extended in a continuous and direct manner to equality of one side of the quadrate c.d, and let that be f.e, and suppose that f.e equals one side of the quadrate c.d. And from e I will draw a right line toward a, and suppose the triangle is orthogonal, for that f is a right angle, and let it be argued following the penultimate of the first, as follows, that the quadrate e.a is as much as the quadrate .e.f and the quadrate fa, but the quadrate e.f is equal to the quadrate c.d, and the quadrate f.a is equal to the quadrate a.b, therefore the quadrate a.e is equal to quadrates a.b and c.d.	¶ Proponantur ergo duo quadrata scilicet .a.b. et .c.d. et sit propositum producere gnomonem circa .a.b. aequalem .c.d. quadrato: protrahatur itaque unum latus quadrati .a.b. ad aequalitatem unius lateris quadrati .c.d. in continuum et directum et sit .f.e. ita quod .f.e. sit aequale uni laterum quadrati .c.d. et ex .e. ducam lineam rectam ad .a. sit ergo triangulus orthogonius quia .f. est angulus rectus arguatur ergo secundum penultimam primi sic: quadratum .e.a. est tantum quantum quadratum .e.f et quadratum .f.a. sed quadratum .e.f. est aequale quadrato .c.d. et quadratum .f.a. est aequale quadrato .a.b. ergo quadratum .a.e. est aequale quadratis .a.b. et .c.d.
Furthermore, e.f.a is a triangle and so sides e.f and f.a are longer than the side a.e according to the 20th of the first. But f.a is equal to a.b by the rationale of a quadrature, and so e.f and f.b are longer than a.e so that whole line, namely, e.b, is greater than a.e, and b.e will thus be resected to equality with a.e at point c so that b.c shall equal a.e, and so the quadrate b.c is equal to the quadrate a.e. But the quadrate a.e, as was first demonstrated, is equal to quadrates a.b and c.d, and so the quadrate b.c is equal to the selfsame, but the quadrate b.c placed over the quadrate a.b is the gnomon that you see, thus, that gnomon is equal to the quadrate c.d, which is that to be proved. The first book ends.	Item .e.f.a. est triangulus ergo .e.f. et .f.a. latera sunt longiora .a.e. latere. secundum .20. primi. sed .f.a. est aequales: .a.b. ratione quadraturae: ergo .e.f. et f.b sunt longiora .a.e. ergo illa totalis linea scilicet .e.b. est maior .a.e. resecetur ergo .b.e. ad aequalitatem .a.e. ad punctum .c. ita quod .b.c. sit aequale .a.e. ergo quadratum .b.c. est aequale. quadrato .a.e. sed quadratum .a.e. ut prius probatum fuit est aequale quadratis .a.b. et .c.d. ergo quadratum .b.c. est aequale eisdem sed quadratum .b.c. addit super quadratum .a.b. gnomonem illum quem vides. ergo gnomo ille est quadrato .c.d. aequalis. quod erat probandum. Explicit liber primus.
The second book begins.	Incipit liber secundus.

[English] Latin	English [Latin][¶]
Every rectangular parallelogram is said to be contained by the two lines subtending a right angle.	Omne paralellogramum rectangulum sub duabus lineis angulum rectum ambientibus dicitur contineri.
¶ A parallelogram is a superficies of equidistant sides.	¶ Paralellogramum est superficies aequidistantium laterum.
¶ A parallelogram having all right angles is rectangular, and is produced from one of its angles from one of its two surrounding sides to the remaining, and so is said to be contained by them.	¶ Paralellogramum rectangulum est habens omnes angulos rectos. et producitur ex uno duorum laterum eius ambientium unum ex suis angulis in reliquum. et ideo sub illis dicitur contineri.
Those parallelograms of every space of every parallelogram which the diameter cuts through the middle are said to be about the selfsame diameter.	Omnis paralellogrami spacii ea quidem quae diameter secat per medium paralellograma circa eandem diametrum consistere dicuntur.
And of those parallelograms which are about the selfsame diameter, any one along with two supplements is called a gnomon.	Eorum vero paralellogramorum quae circa eandem diametrum consistunt quodlibet unum cum supplementis duobus gnomo nominatur.
¶ Some parallelograms are said to be about the diameter and some are the supplements (as was formerly exhibited in demonstration 43 of the first):	¶ Quae paralellograma dicuntur consistere circa diametrum. et quae sunt supplementa: expositum est supra in demonstratione .43. primi.
¶ For suppose the parallelogram is a.b.c.d, whose diameter a.d divides the two lines e.f and g.h (drawn equidistantly from opposite sides of the aforesaid parallelogram) that they are each cutting the other upon diameter a.d at point k, and that will divide the parallelogram into 4 parallelograms. And each of the two parallelograms, which are a.g.e.k and k.f.h.d, that cut the diameter through the middle are said to be about the diameter.	¶ Sit enim paralellogramum .a.b.c.d. cuius diameter .a.d. dividant duae lineae .e.f.g.h. ductae aequidistanter: lateribus oppositis dicti paralellogrami. secantes se super diametrum .a.d. in puncto .k. eritque ipsum paralellogramum divisum in .4. paralellograma. et unumquodque duorum paralellogramorum quae sunt .a.g.e.k. et .k.f.h.d. quae diameter secat per medium dicitur consistere circa diametrum.
The remaining two that do not cut the diameter are called supplements, and these two supplements along with each of the aforesaid parallelograms that are about the diameter construct a certain figure that is called the gnomon, which is the wanting complement of the parallelogram (the one remaining parallelogram that is about the diameter) which, if added on top the diameter will be the total of the composed and will be alike to the whole.	Reliqua duo quae diameter non secat dicuntur supplementa quae duo supplementa cum utroque dictorum paralellogramorum consistentium circa diametrum componunt figuram quandam quae gnomo appellatur cui deest ad complementum paralellogrami paralellogramum unum reliquum circa diametrum consistens: quod si addatur supra diametrum totalis compositi consistet. eritque simile totali.
Hence, the parallelogram may be increased by the addition of the gnomon, though not in the least is it altered; just as Aristotle did state in the Praedicamenta.	Unde paralellogramum addito gnomone quamvis crescat minime tamen alteratur. quemadmodum dixit Aristoteles in praedicamentis.
Proposition 1	Propositio .i.
[English] Latin	English [Latin][¶]
If there were to be two lines, of which one is divided howsoever into parts, then that which will be produced from a drawing of one according to the other will be equal to those rectangles which are produced from drawing indivisible lines at each portion of the partially divided line.	Si fuerint duae lineae quarum una in quodlibet partes dividatur. illud quod ex ductu alterius in alteram fiet. aequum erit his quae ex ductum lineae indivisae in unamquamque partem lineae particulatim divisae rectangula producentur.
¶ To draw a line according to another line is to erect two lines orthogonally over the termini of one of them, equal to the other, and to complete a rectangular surface of equidistant sides, which is said by definition to be contained by those two lines.	¶ Lineam in aliam lineam ducere est supra terminos unius earum duas lineas orthogonaliter alii aequales erigere. et superficiem aequidistantium laterum rectangulum complere quae sub illis duabus lineis per diffinitionem dicitur contineri.
¶ Suppose two lines, a.b and c, of which one, namely a.b, is divided howsoever into parts, which are a.d, and d.e, and e.b. I say that that which is from a drawing of c according to the whole of a.b is equal to those rectangular parallelograms joined together that will be produced from c at a.d, and at d.c, and at e.b.	¶ Sunt duae lineae .a.b. et .c. quarum. una scilicet .a.b. in quodlibet partes dividatur quae sint .a.d. et .d.e. et .e.b. dico quod illud quod sit ex ductu .c. in totum .a.b. aequum est illis paralellogramis rectangulis simul iunctis quae fiunt. ex .c. in .a.d. et in .d.e. et in .e.b.
¶ Over points a and b I shall erect the lines a.f and b.g perpendicular to the line a.b, both of which shall be equal to the line c, and by drawing the line f.g I will complete the rectangular surface a.f.b.g that by definition is produced from c according to a.b, and so is said to be contained by them. I will also extend from points d and e the lines d.h and e.k, equidistant to the sides a.f and b.g, and both of them will be equal to c by the 34th of the first, and each of them is equal to a.f. And so by definition, the rectangle a.d.f.h is produced from c according to a.d, and is said to be contained by them, and the rectangle d.g and e.k from c according to d.e and the rectangle e.k.b.g from c according to e.b, and that these rectangles are jointed together and are equal to the whole rectangle a.f.b.g, it is evident the intent is true.	¶ Super puncta .a.b. erigam lineas .a.f. et .b.g. perpendiculares super lineam .a.b. quarum utraque sit aequalis lineae .c. et complebo rectangulam superficiem .a.f.b.g. ducta linea .f.g. quae per diffinitionem producitur ex .c. in .a.b. et sub illis dicitur contineri. protraham quoque a punctis .d. et .e. lineas .d.h. et .e.k. aequidistantes lateribus .a.f. et .b.g. eritque utraque earum aequalis .c. per .34. primi utraque earum est aequalis .a.f. per diffinitionem igitur rectangulum .a.d.f.h. producitur ex .c. in .a.d. et sub illis dicitur contineri et rectangulum .d.h. et .e.k. ex .c. in .d.e. et rectangulum .e.k.b.g. ex .c. in .e.b. et quia haec rectangula simul iuncta sunt aequalia totali rectangulo .a.f.b.g. patet verum esse propositum.
Proposition 2	Propositio .2.
[English] Latin	English [Latin][¶]
If a line has been divided into parts, that which is produced from a drawing of the whole line according to itself shall equal that which is from a drawing of the same according to all its parts.	Si fuerit linea in partes divisa. illud quod ex ductu totius lineae in seipsam fit: aequum erit his quae ex ductu eiusdem in omnes suas partes:
¶ Suppose the line a.b is divided into a.c and c.d and d.b. I say that that which is produced from a drawing of the whole of a.b according to itself, which would be a.e.b.f, is equal to those which are produced from the whole according to each of the stated parts, which will be plainly evident by drawing c.g and d.h equidistant to a.e and b.f.	¶ Sit linea .a.b. divisa in .a.c. et .c.d. et .d.b. dico quod illud quod fit ex ductu totius .a.b. in se quod sit .a.e.b.f. aequum est his quae fiunt ex ipsa tota in unamquamque dictarum partium quod palam patebit. ductis .c.g. et .d.h. aequidistanter .a.e. et .b.f.
¶ Alternatively, let k be presupposed equal to a.b, and so by the preceding that which is produced from a drawing of k according to the whole of a.b will equal that which is produced from a drawing of k according to all parts of a.b, and because that produced from k according to a.b is as much as from a.b according to itself, and that from k according to all parts of a.b is as much as from a.b according to all parts of the same, and for that k and a.b are equal, it is evident the proposition is true.	¶ Aliter sumatur .k. aequalis .a.b. eritque per praemissam quod fit ex ductu .k. in totam .a.b. aequum ei quod fit ex ductu .k. in omnes partes .a.b. et quia ex .k. in .a.b. tantum fit quantum ex .a.b. in se. et ex .k. in omnes partes .a.b. quantum ex .a.b. in omnes partes eiusdem. propter id quia .k. et .a.b. sunt aequales patet verum esse propositum.
Proposition 3	Propositio .3.
[English] Latin	English [Latin][¶]
If a line has been divided in two parts, that which will be produced from a drawing of the whole according to either part will equal that which is from a drawing of the same part according to itself and the one according to the other.	Si fuerit linea in duas partes divisa illud quod fiet ex ductu totius in alterutram partem aequum erit his quae ex ductu eiusdem partis in seipsam et alterius in alteram.
¶ Suppose the line a.b is divided into a.c and b.c. I say that that which is produced from the whole of a.b according to its part a.c is equal to the quadrate of that same part a.c and that which is produced from that selfsame part a.c according to b.c. Suppose the quadrate of the line a.c, that shall be a.c.d.f, completes the superficies a.b.d.e, and the proposition will be evident.	¶ Sit linea .a.b. divisa in .a.c. et .b.c. dico quod illud quod fit ex tota .a.b. in eius partem .a.c. aequum est quadrato eiusdem .a.c. partis. et ei quod fit ex eadem parte .a.c. in .b.c. fiat quadratum lineae .a.c. quod sit .a.c.d.f. et perficiatur superficies .a.b.d.e. patebitque propositum.
¶ Alternatively, let g be presupposed equal to a.c, and because b.a according to a.c is as much as a.c according to a.b and vise versa, and a.c according to a.b, and to c.b, and to itself, is as much as g according to all the same: then g according to the whole of a.b is as much as g according to a.c and to c.b by the first of this, and the intent is clear, namely, that a.c will be as much according to a.b, as to itself and to c.b, and conversely, a.b according to a.c is as much as a.c according to itself and to c.b, which we mean to demonstrate.	¶ Aliter sumatur .g. aequalis .a.c. et quia .b.a. in .a.c. tantum est quantum .a.c. in .a.b. econverso. et .a.c. in .a.b. et in .c.b. et in seipsam quantum .g. in easdem: At .g. in totam .a.b. quantum in .a.c. et in .c.b. per primam huius patet propositum scilicet quod tamen erit .a.c. in .a.b. quantum in se et .c.b. quare econverso .a.b. in .a.c. quantum .a.c. in se. et in .c.b. quod volumus demonstrare.
Proposition 4	Propositio .4.
[English] Latin	English [Latin][¶]
If a line has been divided into two parts, that which is produced from a drawing of the whole according to itself is equal to that which is from a drawing of each part according to itself, and twice one to the other. From this it is clearly evident that within every quadrate, the two superficies that the diameter divides through the midst are both quadrates.	Si fuerit linea in duas partes divisa illud quod ex ductu totius in seipsam fit: aequum est his quae ex ductu utriusque partis in seipsam et alterius in alteram bis. Ex hoc manifestum est quod in omni quadrato duae superficies quas diameter secat per medium sunt ambae quadratae.
¶ Suppose the line a.b is divided into a.c and b.c. I say that the quadrate of the whole of a.b is equal to the two quadrates of the two lines a.c and b.c, and double which is produced from a drawing of one of them according to the other. I will draw the quadrate of the one part, and c.d.b.e shall be the quadrate of the line c.b to which I will join the gnomon following the guiding directive of the other line, namely, a.c, that I shall make in this way: in the described quadrate I will extend the diameter b.d, and at point a I shall draw a perpendicular to the line a.b, which shall be a.k, and I will extend a.k as far as the diameter b.d until they concur at point f, and from point f I shall draw f.h equidistant to the line a.b. And I will extend f.h as far as b.e until they concur at point g, and I shall draw c.d up until h, and e.d up until k.	¶ Sit linea .a.b. divisa in .a.c. et .b.c. dico quod quadratum totius .a.b. aequum est duobus quadratis duarum linearum .a.c. et .b.c. duplo eius quod fit ex ductu unius earum in alteram: describam quadratum alterius partialium sitque .c.d.b.e. quadratum lineae .c.b. cui adiungam gnomonem secundum ductum directiuum lineae alterius scilicet .a.c. quod faciam hoc modo. in quadrato descripto protraham diametrum .b.d. et a puncto .a. ducam perpendicularem super lineam .a.b. quae sit .a.k. quam .a.k. et diametrum .b.d. producam usquequo concurrant in puncto .f. et a puncto .f. producam f.h. aequidistantem lineae a.b. quam .f.h. et .b.e. producam usque quo concurrant in puncto .g. et producam .c.d. usque ad .h. et .e.d. usque ad .k.
And because the two sides, d.e and e.b, of the triangle d.e.b. are equal by the 5th of the first, the two angles e.d.b and e.b.d will be equal. And because angle e is right by the 32nd of the first, each of them will be half of a right.	Et quia duo latera .d.e. et .e.b. trianguli .d.e.b. sunt aequalia: erunt per .5. primi duo anguli .e.d.b. et .e.b.d. aequales: et quia angulus .e. est rectus erit per .32. primi uterque eorum medietas recti.
By the same rational, both of the two angles, c.d.b and c.b.d, will be half of a right, hence, by the second part of the 29th of the first, each four of the angles, which are h.f.d, and h.d.f, and k.f.d, and k.d.f, will be half of a right, and so by the 6th of the first, f.g and g.b are equal. Similarly as well, f.a and a.b, with like regard, f.h and h.d, and also f.k and k.d, wherefore both of the two surfaces, a.b.g.f and k.d.h.f, are quadrates, and because the whole quadrate a.b.f.g that is the quadrate of the line a.b is from the two quadrates which are about the diameter, which are quadrates of the two lines, a.c and c.b, and from the two supplements, each of which is produced from a.c according to b.c, our proposition is evident.	Eadem ratione uterque duorum angulorum .c.d.b. et .c.b.d. erit medietas recti. quare per secundam partem .29. primi erit unusquisque quatuor angulorum qui sunt .h.f.d. et .h.d.f. et .k.f.d. et .k.d.f. medietas recti ergo per .6. primi .f.g. et .g.b. sunt aequales. similiter quoque .f.a. et .a.b. pari ratione .f.h. et h.d. itemque .f.k. et k.d. quare utraque duarum superficierum .a.b.g.f. et .k.d.h.f. est quadrata et quia totale quadratum a.b.f.g. quod est quadratum lineae .a.b. constat ex duobus quadratis quae consistunt circa diametrum quae sunt quadrata duarum linearum .a.c. et .c.b. et ex duobus supplementis quorum unumquodque producitur ex .a.c. in .b.c. patet propositum nostrum.
¶ Alternatively, let the line a.b be as previously divided into a.c and c.b, and it will be by the 2nd of this that that from the whole of a.b according to itself will equal that which is produced from a.b according to a.c and c.b, but that produced from a.b according a.c is as much as from a.c to itself and a.c to b.c, by the 3rd of this.	¶ Aliter sit linea .a.b. ut prius divisa in .a.c. et .c.b. eritque per .2. huius quod sit ex tota .a.b. in se: aequum ei quod fit ex ipsa in .a.c. et .c.b. sed ex ipsa in .a.c. tantum fit quantum ex .a.c. in se. et ex .a.c in .b.c. per .3. huius.
And further, that produced from the whole of a.b according to b.c will be as much as from c.b according to itself and c.b to a.c by means of the same. Therefore, that produced from the whole of a.b according to itself is equal to that which is produced from a.c to itself and to c.b, and from c.b to itself and to a.c, which is the intent.	Itemque ex ipsa .a.b. tota in .b.c. tantum fit quantum ex .c.b. in se. et ex .c.b. in .a.c. per eandem. ergo quod fit ex tota .a.b. in se aequum est ei quod fit ex .a.c in se et in .c.b et ex .c.b in se. et in .a.c quod est propositum.
But this way the correlation is not evident just as the preceding way is evident, whence the former is more agreeable to the author.	Sed haec via non patet correlarium. sicut via praecedenti patet. unde prima est auctori magis consona.
Proposition 5	Propositio .5.
[English] Latin	English [Latin][¶]
If a right line is cut through into two equal and two unequal parts, the rectangle contained within the inequalities of the entire division along with the quadrate that is defined from that which is between both divisions, is equal to the quadrate that is defined from half the whole line drawn according to itself.	Si linea recta per duo aequalia duoque inaequalia secetur quod sub inaequalibus totius sectionis rectangulum continetur cum eo quadrato quod ab ea quae inter utrasque est sectiones describitur aequum est ei quadrato quod a dimidio totius lineae in se ducto describitur.
¶ Suppose the line a.b is divided by equals at point c and by unequals at point d. I say the quadrate c.b is to be equal to that which is produced from a.d according to d.b and the quadrate c.d.	¶ Sit linea .a.b. divisa per aequalia in puncto .c. et per inaequalia in puncto .d. dico quadratum .c.b. esse aequale ei quod fit ex .a.d. in .d.b. et quadrato .c.d.
¶ I will draw the quadrate c.b, that shall be c.b.f.e, within which I will extend the diameter e.b, and I shall draw d.g equidistant to b.f, which shall cut the diameter e.b at point h, and from point h I will draw out a line equidistant to a.b, by which h.k is cutting the line b.f at point m and the line c.e at point l, and I will draw a.k equidistant to c.e. And by the preceding corollary, each of the two surfaces, l.g and d.m, will be quadrates, and by the 43rd of the first, the two supplements, c.h and h.f, will be equal. Therefore, by adding the quadrate d.m to both, the parallelogram c.m will equal the parallelogram d.f. And because a.l is equal to c.m by the 36th of the first, a.h will equal the gnomon that surround the quadrate of l.g, and so by adding the quadrate l.g to both, a.h along with the quadrate l.g will equal the quadrate e.f, which is the intent.	¶ Describam quadratum .c.b. quod sit .c.b.f.e. in quo protraham diametrum .e.b. et ducam .d.g. aequidistantem .b.f. quae secet diametrum .e.b. in puncto .h. et a puncto .h. educam aequidistantem lineae .a.b. quae sit .h.k. secans lineam .b.f. in puncto .m. et lineam .c.e. in puncto .l. et protraham .a.k. aequidistantem .c.e. eritque per correlarium praemissae utraque duarum superficierum .l.g. et .d.m. quadrata. et per .43. primiduo supplementa c.h. et .h.f. aequalia. rgo addito quadrato .d.m. utrique erit paralellogramum .c.m. aequale paralellogramo .d.f. et quia .a.l. est aequale .c.m. per .36. primi: erit .a.h. aequale gnomoni qui circumstat quadrato .l.g. ergo addito utrique quadrato .l.g. erit .a.h. cum quadrato .l.g. aequale quadrato .e.f. quod est propositum.
Proposition 6	Propositio .6.
[English] Latin	English [Latin][¶]
If a right line is divided in two equal parts but another line is added to it lengthwise, that from a drawing of the whole heretofore composed according to that which has moreover been adjoined, along with that from a drawing of the half according to itself, is equal to that quadrate which is defined by that which is described from the adjoined and the half drawn according to itself.	Si recta linea in duo aequalia dividatur. alia vero ei linea in longum addatur. quod ex ductu totius iam compositae in eam quae iam adiecta est cum eo quod ex ductu dimidiae in seipsam: aequum est ei quadrato quod ab ea quae constat ex adiecta et dimidia in seipsam ducta describitur.
¶ Suppose the line a.b is divided by equals at point c, to which the line b.d is added. I say that the quadrate c.d, that is c.d.e.f, is equal to that which is produced from the whole of a.d according to b.d and the quadrate c.b.	¶ Sit linea .a.b. divisa per aequalia in puncto .c. ei quae addatur linea .b.d. dico quod quadratum .c.d. quod sit .c.d.e.f. aequale est ei quod fit ex tota .a.d. in .b.d. et quadrato .c.b.
I will extend the diameter d.e within the aforementioned quadrate and draw the line b.g equidistant to d.f, which shall cut the diameter d.e at point h, from which point I will extend a line equidistant to a.b, by which h.k is cutting d.f at point m and c.e at point l, and I will extend a.k equidistant to c.l and by the 36th of the first, a.l will equal c.h.	Producam in quadrato praedicto diametrum .d.e. et ducam lineam .b.g. aequidistantem .d.f. quae secet diametrum .d.e. in puncto .h. a quo .h. producam aequidistantem lineae .a.b. quae sit .h.k. secans .d.f. in puncto .m. et .c.e. in puncto .l. et producam .a.k. aequidistantem .c.l. eritque per .36. primi .a.l. aequale .c.h.
Whereas c.h will be equal to h.f by the 43rd of the first and so a.l is equal to h.f, and so by adding c.m to both, a.m will equal the entire gnomon surrounding l.g, wherefore adding l.g to both, a.m along with l.g will equal the total quadrate c.f. And because each of the two superficies, l.g and b.m, are quadrates by the 4th corollary of this, the proposition is evident.	At .c.h. erit aequale .h.f. per .43. primi. quare .a.l. est aequale .h.f. ergo addito .c.m. utrobique erit. a:m. aequalis toti gnomoni circumstanti .l.g. quare .l.g. addito utrobisque erit .a.m. cum .l.g. aequale toti quadrato .c.f. et quia utraque duarum superficierum .l.g. et .b.m. est quadrata: per correlarium .4. huius patet propositum.
Proposition 7	Propositio .7.
[English] Latin	English [Latin][¶]
If a line is divided into two parts, that produced from a drawing of the whole according to itself together with that which is from a drawing of one part according to itself, is equal to twice that which is from a drawing of the whole line according to that same part and from a drawing of the other part according to itself.	Si linea in duas partes dividatur. quod fit ex ductu totius in seipsam cum eo quod est ex ductu alterius partis in seipsam. aequum est quae eis ex ductu totius lineae in eandem partem bis et ex ductu alterius partis in seipsam.
¶ Suppose the line a.b is divided into two parts at point c. I say that the quadrate of the whole of a.b along with the quadrate b.c is equal to twice that produced from a.b according to b.c, with the quadrate a.c. The quadrate of the whole, that is a.b.d.e, is drawn out and the diameter b.d extended. And c.f, equidistant to b.e, is cutting the diameter at point g, and k.g.h is drawn equidistant to a.b, and because the quadrate a.e with the quadrate c.h is so much the quadrate k.f with the two surfaces, a.h and c.e, the proposition is evident.	¶ Sit linea .a.b. divisa in duas partes in puncto .c. dico quod quadratum totius .a.b. cum quadrato .b.c. aequum est ei quod fit ex .a.b. in .b.c. bis cum quadrato .a.c. describatur quadratus totius quod sit .a.b.d.e. et ducatur diametrum .b.d. et .c.f. aequidistans .b.e. secans diametrum in puncto .g. et ducatur .k.g.h. aequidistans .a.b. et quia quadratum .a.e. cum quadrato .c.h. tantum sunt quantum quadratum .k.f. cum duabus superficiebus .a.h. et .g.e. [sic] patet propositum.
Proposition 8	Propositio .8.
[English] Latin	English [Latin][¶]
If a line is divided into two parts and a line equal to one of the divisions is added to it lengthwise, that produced from a drawing of the whole moreover composed according to itself will be equal to four times that which is from a drawing of the original line according to the adjoined, and that from a drawing of the other division according to itself.	Si linea in duas partes dividatur: eique in longum aequalis uni dividentium adiungatur: quod ex ductu totius iam composite in seipsas fiet. aequum erit his quae ex ductu prioris lineae in eam adiectam quater. et ei quod ex ductu alterius dividentis in seipsam.
¶ Suppose a.b is divided at point c, howsoever it may come to pass, to which b.d, equal to c.b, is added. I say that the quadrate of the whole of a.d, that is a.d.e.f, is equal to four times that which is produced from a.b according to b.d, together with the quadrate a.c. And this will be evident by drawing the diameter d.e, and the lines c.g and b.h equidistant to the line d.f, and by cutting the diameter at point k and l (the points through which p.q.k.r and m.n.l.o are drawn equidistant to a.d). And by the 4th corollary of this, each surface, r.g, n.q, and b.m, will be quadrates, and the 4 quadrates dividing the quadrate c.p will be equal. And that the whole gnomon surrounding the quadrate r.g is fourfold that which is from a.b according to b.d, for it is fourfold the surface a.l, the proposition is evident.	¶ Sit .a.b. divisa in puncto .c. qualitercumque contingat: cui addatur .b.d. aequalis .c.b. dico quod quadratum totius .a.d. quod sit .a.d.e.f. est aequale ei quod fit ex .a.b. et .b.d. quater cum quadrato .a.c. hoc autem patebit ducta diametro .d.e. et lineis .c.g. et .b.h. aequidistantibus lineae .d.f. et secantibus diametrum in puncto .k.l. per quae puncta ducantur .p.q.k.r. et .m.n.l.o. aequidistantes .a.d. erit enim per correlarium .4. huius unaquaeque superficierum .r.g.n.q. et .b.m. quadrata: Eruntque .4. quadrata dividentia quadratum .c.p. aequalia et quia totus gnomo circumstans quadrato .r.g. est quadruplus ei quod ex .a.b. in .b.d. quia quadruplus ad superficiem .a.l. patet propositum.
Proposition 9	Propositio .9.
[English] Latin	English [Latin][¶]
If a line is divided in two equal and two unequal parts, that produced from a drawing of the unequal sections according to themselves, taken together, is twice the quadrate of either which are indeed from the halves and the quadrate drawn from that which lies between each section, taken together.	Si linea in duo aequalia duoque inaequalia dividitur: quae fiunt ex ductu inaequalium sectionum in seipsam pariter accepta: duplum sunt utriusque pariter acceptis. quae quidem ex dimidia. eaque quae utrique sectioni interiacet quadratis describuntur.
¶ Suppose the line a.b is divided by equals at c and by unequals at d.	¶ Sit linea .a.b. divisa per aequalia .in.c. et per inaequalia .in.d.
I say that the quadrate a.d and the quadrate d.b joined together is twice the quadrate a.c and the quadrate c.d joined together.	Dico quod quadratum .a.d. et quadratum .d.b. simul iuncta: dupla sunt quadrato .a.c. et quadrato .c.d. simul iunctis.
¶ Over the line a.b I erect the line c.e, perpendicular and equal to both of the lines a.c and c.b, and I draw out e.a and e.b. And by the 32nd of the first, both of the angles a and b and both of the partial angles that are upon e are half a right, and the whole of e is right. And I draw out d.f equidistant to c.e and perpendicular to the line a.b, and both of the angles upon d are right and the angle d.f.b is half right by the 32nd of the first, or by the second part of the 29th of the first, and so by the 6th of the first, d.f and d.b are equal. From point f I draw f.g equidistant to a.b, and by the second part of the 29th of the first, both of the angles upon g will be right and the angle e.f.g half right, and so by the sixth of the same, the sides e.g and g.f are equal. And by the penultimate of the selfsame, that the quadrate e.f is equal to the quadrate e.g and the quadrate g.f, it will be twice the quadrate upon g.f, and hence the quadrate upon c.d.	¶ Super lineam: a.b. erigo lineam .c.e. perpendicularem et aequalem utrique earum linearum .a.c. et .c.b. et produco .e.a. et e.b. eritque per .32. primi uterque angulorum .a. et .b. et uterque angulorun partialium qui sunt ad .e. medietas recti. totusque .e. rectus. et produco .d.f. aequidistantem .c.e. et perpendicularem super lineam .a.b. eritque uterque angulorum .d. rectus: et angulus .d.f.b. medietas recti per .32. primi: sive per secundam partem .29. primi: quare per .6. primi .d.f. et .d.b. sunt aequalia. a puncto .f. duco .f.g. aequidistantem .a.b. eritque per secundam partem .29. primi: uterque angulorum .g. rectus. et angulus .e.f.g. medietas recti quare per sextam eiusdem latera .e.g. et .g.f. sunt aequalia: et quia per penultimam eiusdem quadratum: e.f. est aequale quadrato .e.g. et quadrato .g.f. ipsum erit duplum ad quadratum .g.f. quare ad quadratum .c.d.
¶ And by the same means, the quadrate e.a is equal to the quadrate a.c and the quadrate c.e, and so it shall be twice the quadrate a.c. And because the quadrate a.f is equal to the quadrate e.f and a.e by the same, it shall be twice the quadrate a.c and the quadrate c.d. But the quadrate a.f is also, by means of the same, equal to the quadrate a.d and the quadrate d.f, therefore, the quadrate a.d and the quadrate d.f are double the quadrate a.c and the quadrate c.d, and because the quadrate d.f is equal to the quadrate d.b, the quadrates of the two lines a.d and d.b will be double the quadrates of the two lines that are a.c and c.d, which is the intent.	¶ Itemque per eandem quadratum .e.a. est aequale quadrato .a.c. et quadrato .c.e. ipsum erit duplum ad quadratum .a.c. et quia quadratum .a.f. est aequale quadrato .e.f. et a.e. per eandem ipsum erit duplum ad quadratum .a.c. et ad quadratum .c.d. sed quadratum .a.f. est iterum aequale per eandem quadrato .a.d. et quadrato d.f. ergo quadratum .a.d. et quadratum .d.f. dupla sunt ad quadratum .a.c. et ad quadratum .c.d. et quia quadratum .d.f. est aequale quadrato .d.b. erunt quadrata duarum linearum .a.d. et .d.b. dupla quadratis duarum linearum quae sunt .a.c et .c.d. quod est propositum.
Proposition 10	Propositio .10.
[English] Latin	English [Latin][¶]
If a line is divided into two equal parts and another is added to it lengthwise, it is necessary for the quadrate that is drawn from the whole with the added and the quadrate that is from that which is added, both quadrates taken together, to be twice the quadrate that is from the half and that produced from that which is from the half and the addition, both quadrates taken together.	Si linea in duo aequalia dividatur eique in longum alia addatur: quadratum quod describitur a tota cum addita et quadratum quod ab ea quae addita est. utraque quadrata pariter accepta. ei quadrato quod a dimidia. eique quod ab ea producitur quae ex dimidia adiectaque consistit utrisque quadratis pariter acceptis dupla esse necesse est.
¶ Suppose the line a.b is divided by equals at c and the line b.d is added to it. I say that the two quadrates of the two lines, a.d and b.d, taken together, are twice the two quadrates of the two lines, a.c and c.d, taken together.	¶ Sit linea .a.b. divisa per aequalia in .c. et addita sibi linea .b.d. dico quod duo quadrata duarum linearum .a.d. et .b.d. pariter accepta dupla sunt duobus quadratis duarum linearum .a.c. et .c.d. pariter acceptis.
¶ I erect c.e perpendicular to the line a.b and equal to both of the lines a.c and c.b, and I complete the triangle a.e.b with the lines a.e and e.b, and as in the preceding, both of the angles a and b and each of those that are upon e will be half right by the 32nd of the first, and the whole of e is right. From point e I draw e.f equal and equidistant to c.d and I extend f.d and e.b until they coincide at point g, and I draw the line a.g. And by the last part of the 29th of the first, the angle c.e.f is right. But the angle c.e.b is half right, and so the angle b.e.f is, likewise, half right. And for that by the 33rd of the same f.d is equidistant to c.e, by the 34th of the same the angle f will be right, and thus by the 32nd of the same the angle e.g.f will be half right.	¶ Erigo .c.e. perpendicularem super lineam .a.b. et aequalem utrique linearum .a.c. et .c.b. et perficio triangulum .a.e.b. ductis lineis .a.e. et .e.b. eritque ut in praemissa uterque angulorum .a. et .b. et uterque eorum quae sunt ad .e. medietas recti per .32. primi: totusque .e. est rectus a puncto .e. produco .e.f. aequalem et aequidistantem .c.d. et produco .f.d. et .e.b. quousque concurrunt in puncto .g. et produco lineam .a.g. eritque per ultimam partem .29. primi: angulus .c.e.f rectus sed angulus .c.e.b. est medietas recti. ergo angulus .b.e.f. est similiter medietas recti: et quia per .33. eiusdem .f.d. est aequidistans c.e. erit per .34. eiusdem angulus .f. rectus. ergo per .32. eiusdem. erit angulus .e.g.f. medietas recti.
And by the same means, the angle d.b.g is likewise half right on account that the angle b.d.g is right, and so by the 6th of the same, the two sides, e.f and f.g, are equal and the two sides, d.b and d.g, are likewise equal.	Itemque per eandem angulus .d.b.g. similiter medietas recti: propter id quod angulus .b.d.g. est rectus ergo per .6. eiusdem duo latera .e.f. et .f.g. sunt aequalia.
And so by the penultimate of the same, the quadrate e.g is twice the quadrate e.f and hence, the quadrate c.d.	Itemque duo latera .d.b. et .d.g. sunt aequalia: ergo per penultimam eiusdem quadratum .e.g. duplum est ad quadratum .e.f. quare ad quadratum .c.d.
¶ And by the same means, the quadrate a.e is twice the quadrate a.c, and that the quadrate a.g is by the same equal to the quadrate a.e and e.g, it is likewise equal to quadrate a.d and d.g as well.	¶ Itemque per eandem quadratum .a.e. duplum est ad quadratum .a.c. et quia quadratum .a.g. est per eandem aequale quadrato .a.e. et .e.g. similiter quoque et quadrato .a.d. et .d.g.
Moreover, because the quadrate d.g is equal to the quadrate b.d, the two quadrates of the two lines a.d and b.d, taken together, will be twice the two quadrates of the two lines a.c and c.d, taken together, which is the intent. And this and all the preceding also maintain veracity concerning numbers just as with lines.	At quia quadratum .d.g. est aequale quadrato .b.d. erunt duo quadrata duarum linearum .a.d. et .b.d. pariter accepta dupla duobus quadratis duarum linearum .a.c. et .c.d. pariter acceptis quod est propositum: haec autem et omnes praemissae veritatem habent in numeris sicut in lineis.
Proposition 11	Propositio .11.
[English] Latin	English [Latin][¶]
To divide a given line as follows, so that the rectangle contained within the whole and one portion is equal to the quadrate that is produced from the remaining section.	Datam lineam sic secare. ut quod sub tota et una portione rectangulum continetur: aequum sit ei quod fit ex reliqua sectione quadratum.
¶ Suppose a.b is the given line that we wish to divide as follows, so that produced from the whole and its lesser shall be equal to the quadrate of the greater.	¶ Sit linea data .a.b. quae volumus sic dividere: ut quod ex tota et eius minore producitur aequum sit quadrato maiori.
¶ I draw the quadrate of that, which shall be a.b.c.d, and I divide the side b.d by equals at e and extend a.e, and I extend e.b up until f so that e.f shall be equal to a.e, and from the outer part at b.f I draw a quadrate that is out from the side a.b, resecting a portion equal to b.f, which shall be b.h, and the quadrate described shall be b.f.h.g.	¶ Describo quadratum ipsius quod sit .a.b.c.d. et latus .b.d. divido per aequalia in .e. et produco .a.e. et .e.b. produco usque ad .f. ita quod .e.f. sit aequalis .a.e. et ex .b.f. portione extrinseca: describo quadratum quod ex latere .a.b. resecat portionem aequalem .b.f. quae sit .b.h. et quadratum descriptum sit .b.f.h.g.
I say that a.b is thus divided at point h that that which is produced from the whole of a.b according to its portion h.a is equal to the quadrate h.b. I extend g.h up until k, which will be equidistant to a.c, therefore because the line d.b is divided by equals at e and the line b.f is added to it, by the 6th of this, that produced from d.f according to b.f along with the quadrate e.b will equal the quadrate e.f, and hence the quadrate e.a.	Dico quod .a.b. sic est divisa in puncto .h. quod illud quod fit ex tota .a.b. in eius portionem .h.a. est aequale quadrato h.b. produco .g.h. usque ad .k. quae erit aequidistans .a.c. quia ergo linea .d.b. divisa est per aequalia in .e. et est sibi addita linea .b.f. erit per .6. huius quod fit ex .d.f. in .b.f. cum quadrato .e.b. aequale quadrato .e.f. quare et quadrato .e.a.
And so by the penultimate of the first, the quadrate of the two lines e.b and b.a. Therefore, by removing from both the quadrate of the line e.b, that which is produced from d.f according to b.f, and is the very superficies of d.g, equals the quadrate of the line a.b. Thus by removing from both the parallelogram h.d, the quadrate h.f will equal the parallelogram h.c and because the quadrate h.f is the quadrate of the line h.b and the parallelogram h.c is produced from c.a, which is equal to a.b according to a.h, it is clear the intention has been achieved.	Quare per penultimam primi: quadratis duarum linearum .e.b. et .b.a. ergo dempto ab utrisque quadrato lineae .e.b. erit quod fit ex .d.f. in .b.f. et ipsum est superficies .d.g. aequale quadrato lineae .a.b. ergo dempto ab utrisque paralellogramo .h.d. erit quadratum .h.f. aequale paralellogramo .h.c. et quia quadratum .h.f. est quadratum lineae. h.b. et paralellogramum .h.c. producitur ex .c.a. quae est aequalis .a.b. in .a.h. patet factum esse propositum.
¶ But you should not endeavor to do this with respect to numbers, for it is impossible for a number to thus be divided as is set forth here by the eleventh, just as you will understand by the teaching of the 29th of the sixth.	¶ Ad hoc autem faciendum in numeris non labores: quia impossibile est numerum sic dividi: ut hic undecima proponit sicut scies sexti .29. te docente.
Proposition 12	Propositio .12.
[English] Latin	English [Latin][¶]
Within these triangles that have an obtuse angle, that which subtends the obtuse angle may be as much greater than both the remaining sides that contain the obtuse angle by twice that contained within one of them and that which is joined directly to it at the obtuse angle from a perpendicular deprehended without.	In his triangulis qui obtusum habent angulum: tanto ea quae obtusum subtendit angulum: ambobus reliquis lateribus quae obtusum continent angulum amplius potest. quantum est quod continetur bis sub uno eorum: atque ea quae sibi directe iuncta ad obtusum angulum a perpendiculari extra deprehenditur.
¶ Suppose the triangle a.b.c as having the obtuse angle a. From point c, a line may be drawn perpendicular toward b.a, which by necessity falls outside the triangle a.b.c, otherwise the obtuse angle may be right or less than right by the 16th of the first, and so let c.d be extended perpendicular to the line a.b up until d.	¶ Sit triangulus .a.b.c. habens angulum .a. obtusum. a puncto .c. ducatur linea perpendicularis ad lineam .b.a. quae necessario cadet extra triangulum .a.b.c. alioquin angulus obtusus esset rectus aut minor recto per .16. primi: sit ergo .c.d. perpendicularis super lineam .a.b. productam usque ad .d.
I say that the quadrate of the side b.c that is subtended by the obtuse angle is as much greater to the two quadrates of the two lines surrounding the obtuse angle, a.b and a.c, as twice that which is produced from b.a according to a.d, for the potential of a line is with respect to its quadrate, whence it is said any line may produce as much as is drawn according to itself.	Dico quod quadratum lateris .b.c. quod subtenditur angulo obtuso tanto maius est duabus quadratis duarum linearum a.b. et .a.c. ambientibus ipsum angulum obtusum. quantum est illud quod fit ex .b.a. in .a.d. bis: potentia enim lineae respectu quadrati sui est. unde tantum dicitur posse linea quaelibet quantum in se ducta producit.
For it will be by the 4th of this the quadrate b.d will equal the two quadrates of the two lines b.a and a.d, and twice of that which is produced from b.a according to a.d. And because the quadrate b.c, by the penultimate of the first, is equal to the quadrate b.d and the quadrate d.c, it too will be equal to the quadrate of the three lines b.a, a.d, and d.c, and double of that which is produced from b.a according to a.d. But by the same, the quadrate a.c is equal to the quadrates a.d and d.c, thus the quadrate b.c is equal to the quadrates of the two lines, b.a and c.a, and double of that which is produced from b.a according to a.d, and so b.c may be so much greater than the two lines, b.a and a.c, as is double that which is produced from b.a according to a.d.	Erit enim per .4. huius quadratum .b.d. aequale duobus quadratis duarum linearum .b.a. et .a.d. et duplo eius quod fit ex .b.a. in .a.d. et quia quadratum .b.c. per penultimam primi est aequale quadrato .b.d. et quadrato .d.c. ipsum erit aequale quadratis trium linearum .b.a.a.d. et .d.c. et duplo eius quod fit ex .b.a. in .a.d. sed per eandem quadratum .a.c. est aequale quadratis .a.d. et d.c. ergo quadratum .b.c. est aequale quadratis duarum linearum .b.a. et .c.a. et duplo eius quod fit ex .b.a. in .a.d. quare .b.c. tanto amplius potest duabus lineis .b.a.a.c. quantum est duplum eius. quod fit ex .b.a. in .a.d.
For we have already said that any line is said only able to produce as much when drawn according to itself, which is the intent.	Iam enim diximus quod tantum dicitur posse linea quaelibet quantum in se ducta producit quod est propositum.
Proposition 13	Propositio .13.
[English] Latin	English [Latin][¶]
That which respects an acute angle of every oxygon may be as much less than both the sides containing the acute angle as is twice that contained by one of them, upon which an interior perpendicular stands, and that part of itself which lies between the perpendicular and the acute angle.	Omnis oxigonii tanto ea quae acutum respicit angulum ambobus lateribus angulum acutum continentibus minus potest: quantum est quod bis continetur sub uno eorum cui perpendicularis intra superstat: eaque sui parte: quae perpendiculari anguloque acuto interiacet.
¶ What is proposed here of the subtended side to any acute angle in an oxygonal triangle holds true of a subtended side to any acute angle within every triangle, whether it may be orthogonal, or amblygonal, or oxygonal.	¶ Quod hic proponitur de latere subtenso alicui angulo acuto in triangulo oxigonio veritatem habet de latere subtenso cuilibet angulo acuto in omni triangulo sive fiat orthogonius sive ambligonius sive oxigonius.
¶ Therefore, in the triangle a.b.c, whatever triangle it may be, suppose angle c is acute that if it were oxygonal, a perpendicular may be drawn from each of the angles, a or b, to either base, b.c or a.c, for a perpendicular will always fall within the triangle when it is thus.	¶ Sit ergo in triangulo .a.b.c. quicumque triangulus fuerit. angulus .c. acutus qui si fuerit oxigonius ducatur perpendicularis ab utroque angulorum .a. vel .b. ad utramque basim .b.c. vel .a.c. quia cum sic fuerit semper cadet perpenduclaris intra triangulum.
But if it is amblygonal or orthogonal, a perpendicular drawn from an obtuse or right angle to an opposite side is, as is self-evident, to fall within the triangle and, that I might state simply, since there are two acute angles within every triangle, one of the remaining angles that are a and b will be necessarily acute.	Si autem sit ambligonius aut orthogonius ab angulo obtuso vel recto ducatur perpendicularis ad latus oppositum quam manifestum est cadere intra triangulum: et ut simpliciter dicam cum in omni triangulo sunt duo acuti anguli necessario erit alter reliquorum angulorum qui sunt .a. et .b. acutus.
Then I will draw a perpendicular to that line which lies between two acute angles. Thus, let the angle b of of the triangle a.b.c also be acute, and so I shall draw a perpendicular to b.c (that is a.d) which, as was stated, falls within the triangle, and I say that the quadrate of a.b that is subtended by the acute angle c is so much less than the two quadrates of the two lines a.c and c.b, as is twice that which is produced from b.c according to d.c.	Ducam igitur perpendicularem ad lineam illam quae duobus acutis interiacet. Sit ergo ut trianguli .a.b.c. angulus .b. etiam sit acutus ducam: ergo ad .b.c. perpendicularem qui sit .a.d. quae ut dictum est cadet intra triangulum. dico itaque quod quadratum .a.b. quod subtenditur angulo acuto .c. tanto minus est duobus quadratis duarum linearum .a.c. et .c.b. quantum duplum eius quod fit ex .b.c. in .d.c.
¶ Or rather, I say that the quadrate a.c, which is also subtended by the angle b—which we made acute, whatever angle a may be—is so much less than the two quadrates of the two lines a.b and b.c as is double that which is produced from c.b according to b.d.	¶ Vel dico quod quadratum .a.c. quod etiam subtenditur angulo .b. quae posuimus acutum quicquid fuerit de angulo .a. tanto minus est duobus quadratis duarum linearum .a.b. et .b.c. quantum est duplum eius quod fit ex .c.b. in .b.d.
And by the 7th of this, the quadrate b.c with the quadrate d.c will be equal to that which is twice produced from b.c according to d.c and the quadrate of the other part, namely, b.d. And so by adding each to the quadrate a.d, the quadrate b.c with the quadrates of the two lines a.d and d.c will equal the quadrates of the two lines a.d and d.b and double that which is produced from c.b according to c.d.	Erit enim per .7. huius quadratum .b.c. cum quadrato .d.c. aequale ei quod fit ex .b.c. in .d.c. bis et quadrato alterius partis scilicet .b.d. quare addito utrique quadrato .a.d. erit quadratum .b.c. cum quadratis duarum linearum .a.d. et .d.c. aequale quadratis duarum linearum .a.d. et .d.b. et duplo eius quod fit ex .c.b. in .c.d.
Yet that by the penultimate of the first the quadrate a.c is equal to the quadrates of the two lines a.d and d.c, and the quadrate b.c with the quadrate a.c will equal the quadrates of the two lines a.d and d.b and that which is twice produced from b.c according to c.d. But through the selfsame penultimate of the first, the quadrate a.b is equal to the quadrates of the two lines a.d and b.d, and so the quadrate b.c with the quadrate a.c is equal to the quadrate a.b and that which is twice produced from b.c according to c.d, hence a.b may be so much less as the two sides b.c and a.c as is twice that which is produced from b.c according to c.d, which is the intent. In a similar way you may demonstrate the side a.c that is subtended by the acute angle b may be so much less than the two sides a.b and b.c as is double that which is produced from c.b according to b.d.	At quia per penultimam primi quadratum .a.c. est aequale quadratis duarum linearum .a.d. et .d.c. erit quadratum .b.c. cum quadrato .a.c. aequale quadratis duarum linearum .a.d. et .b.d. et duplo eius quod fit ex .b.c. in .c.d. sed per eandem penultimam primi quadratum .a.b. aequum est quadratis duarum linearum .a.d. et .b.d. ergo quadratum .b.c. cum quadrato .a.c. aequum est quadrato .a.b. et duplo eius quod fit ex .b.c. in .c.d. quare tanto minus potest .a.b. duobus lateribus .b.c. et .a.c. quantum est duplum eius quod fit ex .b.c. in .c.d. quod est propositum. Simili modo probabis latus .a.c. quod subtenditur angulo .b. acuto posse tanto minus duobus lateriubs .a.b. et .b.c. quantum est duplum eius: quod fit ex .c.b in .b.d.
¶ Note, however, by this and the preceding and the penultimate of the first, that by knowing the sides of every triangle, its area is known, and through the auxiliary writings of chord and arc, every angle of it is known.	¶ Notandum autem per hanc et praecedentem et penultimam primi: quod cognitis lateribus omnis trianguli cognoscitur area ipsius et auxiliantibus tabulis de corda et arcu cognoscitur omnis eius angulus.
Proposition 14	Propositio .14.
[English] Latin	English [Latin][¶]
To describe a quadrate equal to a given triangle.	Dato trigono aequum quadratum describere.
¶ Let a be the given triangle to which we want to describe an equal quadrate.	¶ Sit datus trigonus. a cui nos volumus aequum quadratum describere.
I will draw a surface of equidistant sides and of right angles equal to the given triangle according to what the 42nd of the first teaches: and let that superficies be b.c.d.e, of which, if the sides shall be equal we have what we seek, for that will be a quadrate by definition.	Designabo superficiem aequidistantium laterum et rectorum angulorum aequalem trigono dato secundum quod docet .42. primi: sitque superficies illa .b.c.d.e. cuius si latera fuerint aequalia habemus quod quaerimus ipsa enim erit quadrata. per diffinitionem.
But if the sides are unequal, then I will join the smaller of the sides to the larger, according to rectitude, and the line c.f will equal the lesser of the two lines, that is c.e, joined to the larger, that is b.c, according to rectitude.	Si autem latera sint inaequalia tunc adiungam minus ipsorum laterum maiori secundum rectitudinem. sitque linea .c.f. aequalis minori duorum laterum quod est .c.e. adiuncta maiori quod est .b.c. secundum rectitudinem.
I will divide the whole of b.f by equals at point g and make g the center upon the line b.f, and following the length of the line g.b I will draw out the semicircle b.h and will extend the side e.c until it cuts the circumference at point h. I say the quadrate of the line c.h is equal to the given triangle.	Totam .b.f. dividam per aequalia in puncto .g. et facto .g. centro super lineam .b.f. secundum quantitatem lineae .g.b. describam semicirculum .b.h.f. et latus .e.c. producam usquequo secet circumferentiam in puncto .h. dico quod quadratum lineae .c.h. est aequale trigono dato.
I will extend the line g.h and because the line b.f is divided by equals at g and by unequals at c, by the 5th of this, that produced from a drawing of b.c according to c.f along with the quadrate c.g will equal the quadrate g.f, hence the quadrate g.h, and so, by the penultimate of the first, the two quadrates of the two lines g.c and c.h as well. And by removing the quadrate c.g from both, that produced from b.c according to c.f will be equal to the surface b.e, because c.f is equal to c.e, and to the quadrate of the line c.h, therefore the quadrate of the line c.h is equal to the triangle a, which is the intent.	Producam lineam .g.h. et quia linea .b.f. divisa est per aequalia in .g. et per inaequalia in .c. erit per .5. huius quod fit ex ductu .b.c. in .c.f. cum quadrato .c.g. aequale quadrato .g.f. quare et quadrato .g.h. quare per penultimam primi et duobus quadratis duarum linearum .g.c. et .c.h. ergo dempto utrique quadrato .c.g. erit quod fit ex .b.c. in .c.f. quod est aequale superficiei .b.e. eo quod .c.f. est aequale .c.e. quadrato lineae .c.h. quare quadratum lineae .c.h. est aequale trigono .a. quod est propositum:
¶ And note that by this the tetragonal side of anything with one part longer is found, and simply every figure contained by right lines, whatever it may be, inasmuch as we resolve every such figure into triangles. And we will find the tetragonal side of any of those triangles following this doctrine, and by the penultimate of the first we will find one line which can be found within all the tetragonal sides. For example, I now want to find the tetragonal side of the irregular right-lined figure a.b.c.d.e.f, and I divide it into three triangles that are a.b.f, c.d.e, and c.f.e.	¶ Et nota quod per hoc invenitur latus tetragonicum cuiuslibet altera parte longioris et simpliciter omnis figure rectis lineis contente quaecumque fuerit. quoniam omnem figuram talem in triangulos resolvemus et cuiuslibet illorum triangulorum inveniemus tetragonicum latus secundum doctrinam istius. et inveniemus per penultimam primi. lineam unam quae possit in omnia latera tetragonica inventa. verbi gratia volo nunc invenire latus tetragonicum rectilineae figurae irregularis .a.b.c.d.e.f. resolvo eam. in .3. triangulos qui sunt a.b.f.c.d.e. et .c.f.e.
I find as well, following the doctrine of this, the three tetragonal sides of those three triangles that are g.h, h.k, and k.l and I erect h.k perpendicular to g.h and extend g.k, and by the penultimate of the first the quadrate g.k will equal the quadrates of the two lines g.h and h.k. And the third side, k.l, I will erect perpendicular over the line g.k and extend the line g.l, and by the penultimate of the first, g.l will be the tetragonal side of the whole of the proposed right-lined figure.	Invenio quoque secundum doctrinam istius tria latera tetragonica istorum trium triangulorum. qui sunt .g.h.h.k. et .k.l: et erigo .h.k. perpendiculariter super .g.h. et produco .g.k. eritque per penultimam quadratum primi .g.k. aequale quadratis duarum linearum .g.h. et .h.k. et tertium latus .k.l. erigo perpendiculariter super lineam .g.k. et produco lineam .g.l. eritque per penultimam primi .g.l. latus tetragonicum totius figurae rectilineae propositae.
¶ The second book ends.	¶ Explicet liber secundus.
The third book begins.	Incipit liber tertius.

[English] Latin	English [Latin][¶]
Diameters of which are equal, are to be themselves equal circles.	Quorum diametri sunt aequales. ipsos circulos aequales esse.
And the greater of which are greater and the lesser of which are lesser.	Maiores autem quorum maiores et minores quorum minores.
¶ A line is said to contact a circle, that when it touches the circle at either part produced, it does not cut the circle.	¶ Circulum linea contingere dicitur: quae cum circulum tangat in utramque partem eiecta. circulum non secat.
¶ Circles are said to contact one another, that those touching do not cut one another.	¶ Circuli sese contigere dicuntur qui tangentes se invicem non secant.
¶ Right lines in a circle are said to be equally distant from the center when perpendiculars drawn to them from the center are equal.	¶ Rectae lineae in circulo aequaliter distare dicuntur a centro. cum a centro ad ipsas ductae perpendiculares fuerint aequales.
¶ And said to be further distant from the center, upon which the greater perpendicular line falls.	¶ Plus vero distare a centro dicitur. in quam perpendicularis longior cadit.
¶ A right line containing a portion of a circle is called a chord.	¶ Recta linea portionum circuli continens corda nominatur.
¶ But a portion of the circumference is nuncupated an arc.	¶ Portio vero circumferentiae arcus nuncupatur.
¶ And the angle of the portion is that said to be contained by a chord and by an arc.	¶ Angulus autem portionis dicitur qui a corda et arcu continetur.
¶ An angle is said to stand upon the arc, which is contained by two right lines exiting any point of the arc to the termini upon a chord.	¶ Supra arcum angulus consitere dicitur. qui a quolibet puncto arcus ad corde terminos duabus rectis lineis exeuntibus continetur.
¶ A sector of a circle is a figure that is comprised within two lines drawn from the center and beneath the arc that is contained by them.	¶ Sector circuli est figura qui sub duabus a centro ductis lineis et sub arcu qui ab eis conprehenditur continetur.
¶ And the angle that is surrounded by these lines is said to stand upon the center.	¶ Angulus autem qui ab eis lineis ambitur supra centrum consistere dicitur.
¶ Similar portions of circles are said to be by which the angles that stand upon the arc are mutually equal to one another.	¶ Similes circulorum portiones dicuntur in quibus qui supra arcum consistunt anguli sibi invicem sunt aequales.
¶ Likewise are similar arcs, which beget equal angles in the aforesaid manner.	¶ Arcus quoque similes sunt qui aequos angulos praedicto modo suscipiunt.
Proposition 1	Propositio .1.
[English] Latin	English [Latin][¶]
To find the center of a proposed circle, whence it is manifest that when two right lines within the same circle terminate at the circumference, neither one of them shall cut the other orthogonally by equals unless it is to pass over the center.	Circuli propositi centrum invenire. unde manifestum est quod duabus rectis lineis in eodem circulo apud circumferentiam terminatis neutra illarum alteram per aequlia orthogonaliter secat nisi ipsa super centrum transierit.
¶ Let the proposed circle be a.b.c, whose center we wish to find. I draw the line a.c within that circle, howsoever it may happen, which I divide by equals at point d and from which I draw a perpendicular to the line a.c, which I join to the circumference from each side. And suppose e.d.b, which in turn I divide by equals at point f, is what I say to be the center of the circle.	¶ Sit circulus propositus .a.b.c. cuius volumus centrum invenire. duco in ipso circulo lineam .a.c. qualitercumque contingat quam divido per aequalia in puncto .d. a quo duco perpendicularem ad lineam .a.c. quam applico circumferentiae ex utraque parte. sitque .e.d.b. quam rursus divido per aequalia in puncto .f. quem dico esse centrum circuli.
For if it is not, on the contrary it will be elsewhere and so either on the line e.b or outside it.	Si enim non est: erit autem alibi aut in linea .e.b. autem extra.
On the line e.b it is not, for if it were on that, as at point g, the line e.f will be greater than the line e.g; a part, namely, from the whole, which is impossible.	In linea .e.b. non: si enim fuerit in ea ut in puncto .g. erit linea .e.f. maior linea .e.g. pars videlicet toto quod est impossibile.
But if it might be outside the line e.b, as at point h, the lines h.a, h.d, and h.c may be drawn. And because the sides h.d and d.a of the triangle h.d.a are equal to the sides h.d and d.c of the triangle h.d.c, and the base h.a to the base h.c, by the 8th of the first the angle a.d.h will equal the angle c.d.h, whereby both are right, and because the angle a.d.b was also right, a.d.h will equal a.d.b by the 3rd petition of the first, that is to say, a part of the whole, which is impossible. The center of the given circle is thus not elsewhere than at point f, which is the intent.	Quod si fuerit extra lineam .e.b. ut in puncto .h. ducantur lineae .h.a.h.d.h.c. et quia latera .h.d. et .d.a. trianguli .h.d.a sunt aequalia lateribus .h.d. et .d.c. trianguli .h.d.c. et basis .h.a. basi .h.c. erit per .8. primi angulus .a.d.h. aequalis angulo .c.d.h. quare uterque rectus et quia angulus .a.d.b. fuit etiam rectus erit .a.d.h. aequalis .a.d.b. per .3. petitione primi pars videlicet toti quod est impossibile. non est ergo centrum dati circuli alicubi quam in puncto .f. quod est propositum.
Proposition 2	Propositio .2.
[English] Latin	English [Latin][¶]
Upon the circumference of a circle marked by two points, it is necessary for a right line drawn from one to the other to divide the circle.	Super circuli circumferentiam duobus punctis signatis. lineam rectam ductam ab altero ad alterum. circulum secare necesse est.
¶ Suppose that on the circumference of a circle as a.b, whose center is c, there are two points marked, which are a and b. I say that a right line connecting one with the other will divide the circle. Otherwise, it will fall outside the circle, and so suppose, if it were possible, a.e.b is a right line. I will produce the lines c.a and c.b and by the 5th of the first the angle c.a.b will equal c.b.a, and I will furthermore extend the line c.e, which shall cut the circumference at point d. And by the 16th of the first the angle a.e.c will be greater than the angle c.b.e, thus greater than the angle c.a.e, and so by the 18th of the same, the side a.c is greater than the side c.e. And because c.d is equal to c.a, c.d will be greater than c.e, a part from the whole, which is impossible. Therefore, because the line connecting the two points a and b does not traverse outside the circle, it shall divide it, which is the intent.	¶ Sit ut in circumferentia circuli .a.b. cuius centrum sit .c. signata sint duo puncta quae sunt .a. et .b. dico quod linea recta coniungens unum cum cum [sic] altero secabit circulum. Alioquin cadet extra circulum: sitque .a.e.b. linea recta si possibile est: producam lineas .c.a. et .c.b. eruntque per .5. primi: angulus .c.a.b. et .c.b.a. aequales: protraham item lineam .c.e. quae secet circamferentiam in puncto .d. eritque per .16. primi: angulus .a.e.c. maior angulo .c.b.e. quare maior angulo .c.a.e. quare per .18. eiusdem latus .a.c. maius latere .c.e. et quia .c.d. est aequalis .c.a. erit .c.d. maior .c.e. pars toto quod est impossibile: quia ergo linea coniungens duo puncta .a.b. non transibit extra circulum secabit ipsum quod est propositum.

Proposition 3	Propositio .3.
[English] Latin	English [Latin][¶]
If a line is placed within a circle beyond the center and another approaching from the center cuts it by equals, it is to insist upon it orthogonally, and if it shall be upon it orthogonally, it is necessary to divide it by equals.	Si lineam intra circulum praeter centrum collocatam. alia a centro veniens per aequa secat. orthogonaliter super eam insistere. et si in eam orthogonaliter steterit. eam per aequalia dividere necesse est.
¶ Suppose a line as a.b is placed within the circle a.b, of whose center is c, and the line c.d, approaching from the center, divides it by equals. I say that it divides it orthogonally, and conversely, if it divides it orthogonally then clearly it divides it by equals. I shall draw the lines c.a and c.b and shall posit first that it divides it by equals, and so the two sides c.d and d.a of the triangle c.d.a are equal to the two sides c.d and d.b of the triangle c.d.b, and the base c.a to the base c.b, and so by the 8th of the first the angle d of the one is equal to the angle d of the other, thus both are right and so c.d is perpendicular to a.b, which is the intent.	¶ Sit ut lineam .a.b. collocatam intra circulum .a.b. cuius centrum sit .c. linea .c.d. veniens a centro dividat per aequalia: dico quod dividit eam orthogonaliter. et econverso videlicet si dividit eam orthogonaliter dividit eam per aequalia: producam lineas .c.a. et .c.b. et ponam primo quod dividat eam per aequalia: erunt ergo duo latera .c.d. et .d.a. trianguli .c.d.a. aequalia duobus lateribus .c.d. et .d.b. trianguli .c.d.b. et basis .c.a. basi .c.b. ergo per .8. primi: angulus .d. unius est aequalis angulo .d. alterius quare uterque rectus: quare .c.d. est perpendicularis super .a.b. quod est propositum.
¶ I shall posit a second time that c.d is perpendicular to a.b and I will show that it divides a.b by equals, and by this framing both of the angles that are at d are right and so one will equal the other.	¶ Ponam iterum quod .c.d. sit perpendicularis super .a.b. et ostendam quod ipsa dividit .a.b. per aequalia erit enim propter hanc positionem uterque angulorum qui sunt ad .d. rectus quare unus aequalis alteri.
Moreover, that by the 5th of the first the angle c.a.d is equal to the angle c.b.d and the side c.a equals the side c.b, by the 26th of the first the line a.d will equal the line d.b, which is the intent.	At quia per .5. primi angulus .c.a.d. est aequalis angulo .c.b.d. et latus .c.a. aequale lateri .c.b. per .26. primi: eiusdem [sic] erit linea .a.d. aequalis lineae .d.b. quod est propositum.
Proposition 4	Propositio .4.
[English] Latin	English [Latin][¶]
If two lines in turn intersect within a circle and traverse not over the center, it is necessary they are not to be divided by equals.	Si intra circulum duae lineae se invicem secent. et super centrum non transeant. non per aequalia eas secari necesse est.
¶ Suppose that within the circle a.b.c.d, whose center is e, two lines, a.c and b.d, intersect at point f and neither one nor the other traverse through the center. I say that they do not divide themselves equally, so that one is divided by equals by the other.	¶ Sit ut in circulo .a.b.c.d. cuius centrum sit .e. duae lineae .a.c. et .b.d. secent se in puncto .f. et utraque earum vel altera non transeat per centrum. dico quod ipsae non dividunt sese per aequalia: ita quod utraque per aequalia dividatur ab altera.
¶ That if this were possible and it were first supposed as follows so that neither might traverse through the center, from the center e I will draw the line e.f, and by the first of the preceding each of the 4 angles that are a.f.e, e.f.c, b.f.c, and e.f.c, are right, which is impossible; for thus, right would be less than right.	¶ Quod si fuerit hoc possibile: ponatur et sic primo ut neutra transeat per centrum a centro .e. producam lineam .e.f. eritque per primam praemisse unusquisque .4. angulorum: qui sunt .a.f.e.e.f.c.b.f.c. et .e.f.d. rectus quod est impossibile: sic enim rectus esset minor recto.
¶ Then suppose that one of them traverses through the center and the other does not, and say that b.d is passing through the center: I still say that they do not divide themselves by equals. That if it were so, then by the first part of the preceding, when b.d is drawn from the center it should divide a.c by equals and divide it orthogonally, and likewise, a.c should divide b.d orthogonally, and because a.c divides b.d by equals as the adversary posits, it will traverse through the center by a corollary of the first of this, and so both pass through the center, which is contrary the hypothesis.	¶ Sit igitur ut altera eorum transeat per centrum et altera non: sitque .b.d. transiens per centrum adhuc dico quod non dividunt sese per aequalia: quod si sic. tunc per primam partem premisse: cum b.d. ducta a centro dividat .a.c. per aequalia dividat eam orthogonaliter. quare etiam .a.c. dividet .b.d. orthogonaliter: et quia dividit .a.c. ipsam .b.d. per aequalia ut ponit adversarius: ipsa transibit per centrum per correlarium primae huius: quare ambe transeunt per centrum quod est contra ypothesim.
Proposition 5	Propositio .5.
[English] Latin	English [Latin][¶]
The centers of circles cutting one another are to be separate.	Circulorum se invicem secantium centra diversa esse.
¶ Suppose the two circles, a.c.b and a.d.b, are cutting one another over the two points a and b.	¶ Sint duo circuli .a.c.b.a.d.b. secantes se super duo puncta .a. et .b.
I say that the centers of them are different.	Dico quod eorum sunt diversa centrra.
¶ For if they were to have the same center, by definition it would be within a portion shared by each circle, and so suppose that it were at e and the lines e.a and e.f.c were drawn, and so by definition the two lines e.a and e.f would be equal.	¶ Si enim haberent idem centrum ipsum erit per diffinitionem in portione utrique circulo communi: sitque illud e. et ducantur lineae .e.a. et .e.f.c. eruntque per diffinitionem duae lineae .e.a. et .e.f. aequales.
¶ And likewise, by definition the two lines e.a and e.c are equal and so e.f is equal to e.c since both of them are equal to e.a, namely, a part of the whole, which is impossible.	¶ Itemque per diffinitionem duae lineae .e.a. et .e.c. aequales: quare .e.f. est aequales .e.c. cum utraque earum sit aequalis .e.a. pars videlicet toti quod est impossibile.
Proposition 6	Propositio .6.
[English] Latin	English [Latin][¶]
It is necessary the center of circles contacting one another not be the same.	Circulorum sese contingentium non idem centrum esse necesse est
¶ Suppose the two circles, a.b and a.c, are contacting one another at point a.	¶ Sint duo circuli .a.b. et .a.c. contingentes se in puncto .a.
I say that the centers of them are different.	Dico quod eorum sunt diversa centra.
But if they were to have the same center, by definition it would be within the smaller of them when the smaller is placed within the larger, and so let that be d. And suppose the lines d.a and d.b.c are drawn and by definition both of the two lines, d.b and d.c, will be equal to a.d, which is impossible.	Si enim habuerint idem centrum erit per diffinitionem inter minorem eorum cum minor positus fuerit intra maiorem: sitque ipsum .d. et ducantur lineae .d.a. et .d.b.c. erit per diffinionem utraque duarum linearum .d.b. et .d.c. aequalis .a.d. quod est impossibile.
¶ And as to circles contacting one another on the outside, namely of which one is without the other, it is evident by the definition of a center that they do not have the same center.	¶ De circulis autem se contingentibus extra quorum scilicet unus est extra alterum: manifestum est per diffinitionem centri quod ipsi non habent idem centrum.
Proposition 7	Propositio .7.
[English] Latin	English [Latin][¶]
If upon the diameter of a circle a point is marked beyond the center and from it a great many lines are drawn to the circumference, that which passes over the center will be longest of them all, and that which completes the diameter shall be the shortest, and those nearer the center shall be longer than the rest. But the farther from the center they are then the shorter they are to be, and moreover, it is necessary that the two shortest equidistant collateral lines are to be equal.	Si in diametro circuli punctus praeter centrum signetur: et ab eo ad circumferentiam lineae plurimae ducantur quae super centrum transierit omnium erit longissima. quae vero dyametrum perficiet omnium erit brevissima. quae autem centro proximae ceteris longiores. Quanto vero a centro remotiores tanto breviores esse convenient. Duas quoque aequidistantes lineae brevissimae collaterales aequales esse necesse est.
¶ Suppose that on the diameter a.f of the circle a.b.c, whose center is h, the point k is marked beyond the center from which very many lines are drawn, which are k.a, k.b, k.c, k.d, k.e, k.f, and k.g, to the circumference, and a.k passes through the center h and k.f is the complement of the diameter, and suppose that k.e and k.g are equidistant to k.f, that is to say, that the angle e.k.f shall be equal to the angle f.k.g. I say that k.a is longest of them all and k.f is shortest of them all, and the others are as much longer as they are nearer to the center, so that k.b is longer than k.c and k.c is longer than k.d and k.d is longer than k.e, and k.e and k.g are equal. And because upon the triangle b.k.h the two sides b.h and h.k are, by the 20th of the first, greater than the side b.k and they are equal to the line a.k, a.k will be greater than b.k and by the same rationale, greater than all the rest, and this is the first.	¶ Sit ut in diametro .a.f. circuli .a.b.c. cuius centrum sit .h. sit signatus punctus .k. praeter centrum a quo ducantur plurimae lineae quae sunt .k.a.k.b.k.c.k.d.k.e.k.f.k.g. ad circumferentiam: et transeat .a.k. per centrum .h. et .k.f. sit complementum dyametri: sitque ut .k.e. et .k.g. aequidistant a .k.f. hoc est dicere ut angulus .e.k.f. sit aequalis angulo .f.k.g. dico quod .k.a. est omnium longissima. et .k.f. omnium brevissima: aliae vero tanto longiores quanto centro propinquiores: ut .k.b. est longior .k.c. et k.c. est longior .k.d. et .k.d. longior .k.e. et .k.e. et k.g. sunt aequales: quia enim in triangulo .b.k.h. duo latera .b.h. et .h.k. per .20. primi: sunt maiora latere .b.k. et ipsa sunt aequalia lineae .a.k. erit .a.k. maior b.k. et eadem ratione maior omnibus aliis et hoc est primum.
¶ And too, because upon the triangle e.h.k the two sides h.k and k.e, by the same, are greater than the side h.e, which is equal to the line h.f, they will be greater than the line h.f. Thus, by removing the common line, which is h.k, k.e will remain greater than k.f and by the same rationale, it will be greater than any of the rest, and this is the second.	¶ Itemque quia in triangulo .e.h.k. duo latera .h.k. et .k.e. per eandem sunt maiora latere .h.e. quod est aequale lineae .h.f. ipsa erunt maiora linea .h.f. ergo dempta communi linea quae est .h.k. remanebit .k.e. maior k.f. eandem ratione quaelibet aliarum erit maior ipsa et hoc est secundum.
¶ And too, because the two sides b.h and h.k of the triangle b.h.k are equal to the two sides c.h and h.k of the triangle c.h.k, and the angle b.h.k is greater than the angle c.h.k, by the 24th of the first the base b.k will be greater than the base k.c, and by the same rationale, k.c will be greater than k.d and k.d greater than k.e, and this is the third.	¶ Itemque quia duo latera .b.h. et .h.k. trianguli .b.h.k. sunt aequalia duobus lateribus .c.h. et .h.k. trianguli .c.h.k. et angulus .b.h.k. est maior angulo .c.h.k. erit per .24. primi basis .b.k. maior basi .k.c. eadem ratione .k.c. maior erit .k.d. et .k.d. maior .k.e. et hoc est tertium.
¶ But if the two lines k.g and k.e are not equal, one will be greater. And let k.g, from which I shall take k.l, be equal to k.e, and I will extend h.l until it cuts the circumference at point m, and that by the hypothesis the angle g.k.f is equal to the angle f.k.e, by the 13th of the first the angle l.k.h will equal the angle e.k.h and the two sides l.k and k.h of the triangle l.k.h are equal to the two sides e.k and k.h of the triangle e.k.h, and so by the 4th of the first, the base h.l is equal to the base h.e and because h.m is equal to h.e, h.m will be equal to h.l, which is impossible. The two lines, k.g and k.e, are therefore equal, which is our fourth intention.	¶ Quod si duae lineae .k.g. et .k.e. non sunt aequales erit altera maior: sitque .k.g. de qui sumam .k.l. aequalem .k.e. et producam .h.l. quousque secet circumferentiam in puncto .m. et quia per ypothesim angulus .g.k.f. est aequalis angulo .f.k.e. erit per .13. primi: angulus .l.k.h. aequalis angulo .e.k.h. et duo latera .l.k. et .k.h. trianguli .l.k.h. sunt aequalia duobus lateribus .e.k. et .k.h. trianguli .e.k.h. ergo per.4. primi basis .h.l. est aequalis basi .h.e. et quia .h.m. est aequalis .h.e. erit .h.m. aequalis .h.l. quod est impossibile. sunt ergo duae lineae .k.g. et .k.e. aequales quod est nostrum propositum .quartum.
Proposition 8	Propositio .8.
[English] Latin	English [Latin][¶]
If from a designated point without a circle a great many lines are drawn to the circumference by means of intersecting the circle, that which passes over the center will be longest of them all and those more near the center will be longer than those more remote. But of the parts of the line applied to the circumference without, that which is directly adjacent to the diameter is the shortest of all and those nearer that are shorter than those more remote, and the two which are equally near either side of the shortest line, are equal.	Si extra circulum puncto signato ab eo ad circumferentiam lineae plurimae ducantur circulum secando. quae super centrum transierit omnium erit longissima. Centro autem propinquiores ceteris remotioribus longiores. Lineae vero partiales ad circumferentiam extrinsecus applicatae: ea quidem quae diametro in directum adiacet omnium est minima. eique propinquiores remotioribus breviores. Duae vero quae lineae brevissimae utrumque aequae propinquant aequales sunt.
¶ Suppose that from the assigned point a, without the circle b.c.d of whose center is n, many lines are drawn to the circumference by intersecting the circle, which are a.k.n.b, a.h.c, a.g.d, and a.f.e.	¶ Sit ut in puncto .a. assignato extra circulum .b.c.d. cuius centrum sit .n. ducantur plurimae lineae ad circumferentiam secando circulum quae sint .a.k.n.b.a.h.c.a.g.d. et .a.f.e.
I say because a.b passes through the center it will be longest of all, and that a.c is longer than a.d and a.d longer than a.e, and that a.k is shorter than all without, and that a.h is shorter than a.g and a.g shorter than a.f. And I say that if a.l were drawn, that because it and a.h are equally distant from a.k—this is to say that angle k.a.h would equal angle l.a.k—that they will be equal.	Dico quod .a.b. transiens per centrum omnium erit longissima. et quod .a.c. est maior .a.d. et .a.d. maior .a.e. et quod .a.k. est omnium brevissima extrinsecarum: et quod .a.h. est minor .a.g. et .a.g. minor .a.f. et dico quod si ducatur .a.l. ita quod ipsa et .a.h. aequaliter distent ab .a.k. hoc est quod angulus .k.a.h. sit aequalis angulo .l.a.k. ipsae erunt aequales.
¶ So I will draw from the center n the lines n.c, n.d, n.e, n.f, n.g, and n.h, and by the 20th of the first the two sides a.n and n.c of the triangle a.n.c are greater than a.c and because they are equal to the line a.b, a.b will be longer than a.c and by the same rationale, longer than all the rest, which is first. And because the two sides a.n and n.c of the triangle a.n.c are equal to the two sides a.n and n.d of the triangle a.n.d and the angle a.n.c is greater than the angle a.n.d, by the 24th of the first the base a.c will be greater than the base a.d and by the same rationale, a.d will be greater than a.e, which is second.	¶ Producam enim a centro .n. lineas .n.c.n.d.n.e.n.f.n.g. et .n.h. eruntque per .20. primi duo latera .a.n. et .n.c. trianguli .a.n.c. maiora .a.c. et quia ipsa sunt aequalia lineae .a.b. erit .a.b. maior .a.c. eadem ratione erit maior omnibus aliis quod est primum. et quia duo latera .a.n. et .n.c. trianguli .a.n.c. sunt aequalia duobus lateribus .a.n. et .n.d. trianguli .a.n.d. et angulus .a.n.c. est maior angulo .a.n.d. erit per .24. primi: basis .a.c. maior basi .a.d. et eadem ratione erit .a.d. maior .a.e. quod est secundum.
¶ And further, because the two sides a.h and n.h upon the triangle a.n.h are greater than a.n by the 20th of the first, and h.n is equal to n.k, through common knowledge a.h will be greater than a.k and by the same rationale, any of those applied without will be longer than a.k, which is third.	¶ Itemque quia in triangulo .a.n.h. duo latera .a.h. et .n.h. sunt maiora .a.n. per .20. primi. et .h.n. est aequalis .n.k. erit per communem scientiam .a.h. maior .a.k. eadem ratione quaelibet extrinsecus applicatarum maior erit .a.k. quod est tertium.
¶ Likewise, that by the 21st of the first the two lines a.h and h.n are lesser than the two lines a.g and g.n and h.n is equal to g.n, through common knowledge a.g will be longer than a.h and by the same rationale, a.f will be longer than a.g, which is fourth.	¶ Item quia per .21. primi: duae lineae .a.h. et .h.n. sunt minores duabus lineis .a.g. et .g.n. et .h.n. est aequalis .g.n. erit per commmunem scientiam .a.g. maior .a.h. eadem ratione erit .a.f. maior .a.g. quod est quartum.
¶ But if a.l were not equal to a.h when it is equally distant from a.k, then one will be greater: and let it be a.l. I will thus suppose a.m equal to a.h and produce n.o.m, and so because the two sides m.a and a.n of the triangle m.a.n are equal to the two sides h.a and a.n of the triangle h.a.n and the angle m.a.n is equal to the angle h.a.n, by the 4th of the first the base m.n is equal to the base n.h and because m.o is equal to n.h, n.o will be equal to n.m, part of the whole, which is impossible, and this is the fifth.	¶ Quod si .a.l. non sit aequalis .a.h. cum ipse sint aequaliter distantes ab .a.k. erit altera maior: sitque .a.l. ponam ergo .a.m. aequalem .a.h. et producam .n.o.m. quia ergo duo latera .m.a. et .a.n. trianguli .m.a.n. sunt aequalia duobus lateribus .h.a. et .a.n. trianguli .h.a.n. et angulus .m.a.n. est aequalis angulo .h.a.n. erit per .4. primi: basis .m.n. aequalis basi .n.h. et quia .m.o. est aequalis .n.h. erit .n.o. aequalis .n.m. pars videlicet toti quod est impossibile et hoc est quintum.
Proposition 9	Propositio .9.
[English] Latin	English [Latin][¶]
If more than two lines drawn to the circumference from a designated point within a circle shall be equal, it is necessary for that point to be the center of the circle.	Si intra circulum puncto signato. ab eo plures quam duae lineae ductae ad circumferentiam fuerint aequales. punctum illud centrum circuli esse necesse est.
¶ Suppose that from the point designated a within the circle b.c.d, there are 3 lines drawn, a.b, a.c, and a.d, to the circumference, which I suppose to be equal. I say point a is the center of the circle.	¶ Sit ut a puncto .a. signato intra circulum .b.c.d. ductae sint .3. lineae .a.b.a.c.a.d. ad circumferentiam quas pono esse aequales dico punctum .a. esse centrum circuli.
So I will draw the two lines c.b and d.c and divide both of them by equals, that is, c.b at point e and d.c at point f, and I will draw e.a and f.a and apply them to the circumference from both ends, and by the 8th of the first each of the angles that are against e equal one another, and so by the 13th they will both be right.	Producam enim duas lineas .c.b. et .d.c. et dividam utramque earum per aequalia .c.b. quidem in puncto .e. et .d.c. in puncto .f. et producam .e.a. et .f.a. quas applico circumferentiae ex utraque parte. eritque per .8. primi uterque angulorum qui sunt ad .e. aequalis alteri. igitur per .13. uterque erit rectus.
And likewise by the same, both of the angles that are against f are right: and so by a corollary of the first of this, because a.e divides c.b by equals and orthogonally, it passes through the center. Similarly, a.f also passes through the center, whereby it divides d.c by equals and orthogonally, and so a is the center, which is the intent.	Similiter quoque per eandem uterque angulorum qui sunt ad .f. rectus: ergo per correlarium primae huius. quia .a.e. dividit .c.b. per aequalia et orthogonaliter ipsa transit per centrum. similiter quoque .a.f. transit per centrum. qua dividit .d.c. per aequalia et orthogonaliter. quare .a. est centrum quod est propositum.
Proposition 10	Propositio .10.
[English] Latin	English [Latin][¶]
If a circle intersects a circle, it is necessary to cut only at two points.	Si circulus circulum secet. in duobus tantum locis secare necesse est.
¶ Suppose it is possible that two circles were cutting one another at more than two points, and over the 3 points, a, b, and c, I will draw the lines a.b and a.c that I will divide by equals at points d and e. And I will draw from point e the line e.f, perpendicular to the line a.c, and from point d the line d.f, perpendicular to the line a.b, and the two lines e.f and d.f intersect at point f, and by a corollary of the first of this, point f is the center of the circle of both, which is impossible (by the 5th of this).	¶ Sint si possibile est duo circuli secantes se in pluribus quam in duobus locis super .3. puncta .a.b.c. producam lineas .a.b. et .a.c. quas dividam per aequalia in punctis .d. et .e. et producam a puncto .e. lineam .e.f. perpendicularem super lineam .a.c. et a puncto .d. lineam .d.f. perpendicularem super lineam .a.b. et secent se duae lineae .e.f. et .d.f. in puncto .f. eritque per correlarium primae huius punctum .f. centrum circuli utriusque quod est impossibile. per .5. huius.
Proposition 11	Propositio .11.
[English] Latin	English [Latin][¶]
If a circle contacts a circle and a line passes through the centers of them, it is necessary that it be applied to the point of their contact.	Si circulus circulum contingat. lineaque per centra eorum transeat. ad punctum contactus earum applicari necesse est.
¶ If a line were passing through the centers of the two circles c.e and d.e, which are contacting one another either from within or without, and it does not extend to the point of contact whilst it cuts the circumference of them each, then suppose a is the center of the circle e.d and b is the center of the circle e.c, and say the right line a.b.c.d is drawn so as to be cutting the circumference of both, and lines are drawn from point e, that is the point of contact, to the centers, which are e.a and e.b. And from the contact to an interior circle, by the 20th of the first, the two lines e.b and b.a are longer than e.a, and so they are longer than a.d, for a is the center of the circle e.d. And because b.c is equal to e.b, for b is the center of the circle e.c, c.a will be longer than a.d, which is impossible.	¶ Si enim linea transiens per centra duorum circulorum .c.e. et .d.e. sese contingentium intra vel extra. non vadit ad locum contactus secet circumferentiam utriusque: sitque .a. centrum circuli .e.d. et .b. centrum circuli .e.c. et ducatur linea recta .a.b.c.d. secans circumferentiam utriusque: et ducantur lineae a puncto .e. qui sit locus contactus ad centra quae sint .e.a.e.b. eruntque in contactu interiori. per .20. primi duae lineae .e.b. et .b.a. longiores .e.a. quare longiores .a.d. est enim .a. centrum circuli .e.d. et quoniam .b.c. est aequalis .e.b. quam .b. est centrum circuli .e.c. erit .c.a. longior .a.d. quod est impossibile.
¶ And from the contact to an exterior circle, the two lines a.e and e.b are longer than a.b thus a.d and c.b are greater than the whole of a.b, which is false.	¶ In contactu vero exteriori erunt duae lineae .a.e. et .e.b. longiores .a.b. quare .a.d. et .c.b. maius erunt quod tota .a.b. quod est falsum.
Proposition 12	Propositio .12.
[English] Latin	English [Latin][¶]
If a circle contacts a circle either within or without, it is necessary for it to contact at only one point.	Si circulus circulum contingat sive intrinsecus sive extrinsecus. in uno tantum loco contigere necesse est.
¶ If it were possible for a circle to contact a circle at two points, within or without, then suppose that the circle a.b.e might contact the circle a.b.c.d interiorly at the two points a and b, and the circle c.d.f contacts it exteriorly at the two points c and d.	¶ Si enim fuerit possibile. ut circulus circulum contingat in duobus locis intra vel extra contingat circulum .a.b.c.d. circulus .a.b.e. interius in duobus punctis .a.b. vel exterius circulus .c.d.f. in duobus punctis .c.d.
Thus, when we shall draw a right line from a to b, if it falls without the interior circle a.b.e it falls contrary to the second of this.	Cum ergo ducemus lineam rectam ab .a. ad .b. si ipsa cadat extra circulum .a.b.e. interiorem accidet contrarium secundae huius.
But if it falls within, when we divide that line by equals and draw from the point of division a line perpendicular to it, and it is applied to the circumference from either side, it will pass through the center of both circles and so falls contrary to the premise.	Quid si ipsa cadat intra ipsum: cum diviserimus ipsam per aequalia et eduxerimus a puncto divisionis perpendicularem ad ipsam. fueritque applicata circumferentiae ex utraque partem ipsa transibit per centrum amborum circulorum. quare accidet contrarium praemissae.
¶ And as to the circle contacting exteriorly at points c and d, if we draw a right line from point c to point d, it is necessary for it to fall contrary to the second of this, wherefore both are impossible.	¶ In circulo vero contingente exterius in punctis .c.d. si ducamus lineam rectam a puncto .c. ad punctum .d. necesse est accidere contrarium secundae huius. quare utrumque impossibile.
Proposition 13	Propositio .13.
[English] Latin	English [Latin][¶]
If right lines within a circle are equal, they are to be equidistant from the center. And if they are equidistant from the center it is necessary they are to be equal.	Rectae lineae in circulo si fuerint aequales eas a centro aequidistare. et si a centro aequidistiterint aequales esse necesse est.
¶ Suppose that within the circle a.b.c.d, whose center is e, the two lines a.d and b.c are equal. I say that they are equidistant from the center, and vise versa.	¶ Sit ut in circulo .a.b.c.d. cuius centrum sit .e. duae lineae .a.b. [sic] et .c.d. [sic] sint aequales. dico quod ipsae aequidistant a centro et econverso.
For if the lines e.f and e.g are drawn from the center e perpendicular to a.d and b.c, by the 2nd part of the third of this a.d will be divided by equals at f, and b.c at g, and so because the two sides e.d and d.a of the triangle e.d.a are equal to the two sides e.c and c.b of the triangle e.c.b, and the base e.a to the base e.b, by the 8th of the first the angle d will equal the angle c. And because the two sides e.d and d.f of the triangle e.d.f are equal to the two sides e.c and c.g of the triangle e.c.g, and since d.f is equal to c.g, and because a.d in whole is posited equal to b.c, and angle d is equal to angle c, then by the 4th of the first the base e.f will equal the base e.g and as they are approaching perpendicularly from the center then by definition (or by the 4th of this) it is clear they are equally distant from the center.	Producantur enim a centro .e. lineae .e.f. et .e.g. perpendiculares ad .a.d. et .b.c. eritque per .2. partem tertiae huius .a.d. divisa per aequalia. in .f. et .b.c. in .g. quia ergo duo latera .e.d. et .d.a. trianguli .e.d.a. sunt aequalia duobus lateribus .e.c. et .c.b. trianguli .e.c.b. et basis .e.a. basi .e.b. erit per .8. primi angulus .d. aequalis angulo .c. et quia duo latera .e.d. et .d.f. trianguli .e.d.f. sunt aequalia duobus lateribus .e.c. et .c.g. trianguli .e.c.g. Nam .d.f. est aequalis .c.g. eo quod tota .a.d. posita est aequalis .b.c. et angulus .d. est aequalis angulo .c. erit per .4. primi basis .e.f. aequalis basi .e.g. et quia istae sunt perpendiculares venientes ad eas a centro patet per diffinitionem: sive .4. huius ipsas aequaliter distare a centro.
¶ And the same in turn:	¶ Aliter idem.
For that the quadrate e.d, by the penultimate of the first, is worth the quadrates of the two lines e.f and f.d and the quadrate e.c [equals] the quadrates of the two lines that are e.g and c.g, and because the quadrate d.e is equal to the quadrate e.c and quadrate d.f [equals] quadrate g.c, the quadrate e.f will equal the quadrate e.g, thus e.f is equal to e.g and so the same shall be evident.	Quadratum enim .e.d. per penultimam primi valet quadrata duarum linearum .e.f. et .f.d. et quadratum .e.c. quadrata duarum linearum quae sunt .e.g. et .c.g. et quia quadratum .d.e. est aequale quadrato .e.c. et quadratum .d.f. quadrato .g.c. erit quadratum .e.f. aequale quadrato .e.g. quare .e.f. est aequale .e.g. sitque patet idem.
Suppose then e.f is equal to e.g, which is to distance them equally from the center. Then I say that a.d is equal to b.c because the quadrates of the two lines e.d and e.c, being equal, removed from the quadrates of the two lines e.f and e.g, being equal, means by the penultimate of the first the quadrates of the two lines f.d and g.c remain, thus through common knowledge it is necessary that they are to be equal, and so f.d is equal to g.c and so double f.d, which is a.d, is equal to double g.c, which is b.c, and this is the second part of the proposition.	Sit ergo .e.f. aequalis .e.g. quod est eas aequaliter distare a centro. dico tunc quod .a.d. est aequalis .b.c.d.e. [sic] quadratis enim duarum linearum .e.d. et .e.c. aequalibus demptis quadratis duarum linearum .e.f. et .e.g. aequalibus remanent per penultimam primi quadrata duarum linearum .f.d. et .g.c. quare per communem scientiam necesse est esse aequalia: quare .f.d. est aequalis .g.c. ergo duplum .f.d. quod est .a.d. est aequale duplo .g.c. quod est .b.c. et haec est secundam pars propositi.
Proposition 14	Propositio .14.
[English] Latin	English [Latin][¶]
If a great many right lines fall within a circle, it is necessary the diameter of it be the longest of all, and those nearer to that to be longer than those more remote.	Si intra circulum plurimae rectae lineae ceciderint diametrum eius omnium longissimam. eique propinquiores remotioribus longiores esse necesse est.
¶ Suppose that in the circle a.b.c, whose center is e, many lines do fall, which are a.b, a.c, a.d, f.g, h.k, and a.e.d shall be the diameter. I say that is to be longest, and the rest are as long as they are near to it. Let lines then be drawn from the center e to the extremities of them all, which are e.b, e.c, e.f, e.h, and e.k, and by the 20th of the first, the two sides e.f and e.g of the triangle e.f.g are longer than f.g, and because they are equal to a.d, a.d will be greater than f.g. By the same rationale it will be greater than a.c for a.e and e.c are greater than a.c, and equal to a.d, thus a.d is greater than a.c. And again in this way, is it greater than h.k and also greater than a.b.	¶ Sit ut in circulo .a.b.c. cuius centrum .e. cadant plurimae lineae quae sint .a.b.a.c.a.d.f.g.h.k. sitque .a.e.d. diameter. dico ipsam esse longissimam et alias tanto maiores quanto sunt ipsi propinquiores. ducantur enim a centro .e. lineae ad extremitates omnium quae sint .e.b.e.c.e.f.e.h. et .e.k. eruntque per .20. primi duo latera .e.f. et .e.g. trianguli .e.f.g. longiora .f.g. et quia ipsa sunt aequalia .a.d. erit .a.d. maior .f.g. eadem ratione maior erit quam .a.c. quia .a.e. et .e.c. sunt maiora .a.c. et aequalia .a.d. ergo .a.d. maior est .a.c. sic quoque est maior .h.k. et maior etiam quam .a.b.
Moreover, that f.g is greater than h.k, and a.c than a.b, is evident that by the 24th of the first, when the two sides f.e and e.g of the triangle f.e.g are equal to the two sides h.e and e.k of the triangle h.e.k and the angle f.e.g is greater than the angle h.e.k, the base f.g will be greater than the base h.k.	Quod autem .f.g. sit maior .h.k. et .a.c.a.b. patet. quia per .24. primi cum duo latera .f.e. et .e.g. trianguli .f.e.g. sint aequalia duobus lateribus .h.e. et .e.k. trianguli .h.e.k. et angulus .f.e.g. maior angulo .h.e.k. erit basis .f.g. maior basi .h.k.
Similarly as well, because a.e and e.c are equal to a.e and e.b and the angle a.e.c is greater than the angle a.e.b, the base a.c will be greater than the base a.b and thus is the intent.	Similiter quoque quia .a.e. et .e.c. sunt aequalia .a.e. et .e.b. et angulus .a.e.c. maior angulo .a.e.b. erit basis .a.c. maior basi .a.b. et sic est propositium.
Proposition 15	Propositio .15.
[English] Latin	English [Latin][¶]
If a right line is drawn orthogonally at either terminus of the diameter of any circle, it is necessary for it to fall without the circle, and it is impossible for any other right line to be contained between it and the circle. Moreover, the angle contained by it and the circumference is to be the narrowest of all acute angles, and it is necessary for the angle contained interiorly by the diameter and circumference to be the widest of all acute angles, whence it is further manifest that every right line drawn orthogonally at the end of the diameter of any circle is to contact that circle.	Si ab altero terminorum diametri cuiuslibet circuli orthogonaliter linea recta ducatur: extra circulum eam cadere necesse est. Atque inter illam et circulum aliam lineam rectam capi impossibile est. Angulum autem ab illa et circumferentia contentum. omnium acutorum angulorum esse angustislimum. Angulum vero intrinsecum a diametro et circumferentia contentum omnium angulorum acutorum esse amplissimum necesse est. Unde etiam manifestum est omnem lineam rectam a termio diametri cuiuslibet circuli orthogonaliter ductam circulum ipsum contingere.
¶ Suppose that an orthogonal line is drawn at a terminus of the diameter a.c, of the circle a.b.c, whose center is d. I say that it falls without the circle, and that no other right line may be interposed between that line and the circumference, and that the angle which it and the circumference contains is smaller than every right-lined angle, namely, that contained by two right lines, and that the angle contained by the diameter and circumference is greater than every acute right-lined angle.	¶ Sit ut a termino a diametri .a.c. circuli .a.b.c. cuius centrum .d. ducatur linea orthogonaliter: dico quod ipsa cadit extra circulum. et quia inter lineam illam et circumferentiam nulla alia recta linea intercipitur: et quod angulus quem ipsa et circumferentia continet est minor omni angulo rectilineo qui videlicet a duabus rectis lineis continetur. et quod angulus contentus a diametro et circumferentia est maior omni angulo rectilineo acuto.
¶ So if a line is drawn at a, orthogonally over the line a.c, and it were able to fall within the circle, suppose that line is a.b, and the line d.b then is drawn. And by the 5th of the first the angle d.a.b equals the angle d.b.a, and that by the hypothesis the angle d.b.a is right, the triangle a.d.b will have two right angles, which is impossible by the 32nd of the first.	¶ Si enim linea ducta. ab.a orthogonaliter super .a.c. lineam: potest cadere infra circulum. sit illa linea .a.b. et ducatur linea .d.b. eritque per .5. primi angulus .d.a.b. aequalis angulo .d.b.a. et quia angulus .d.b.a. est rectus per ypothesim. habebit triangulus .a.b.d. duos angulos rectos quod est impossibile. per .32. primi:
¶ Therefore, it will fall without and shall be as a.e. But if a right line were able to be interposed between it and the circumference, then suppose that to be a.f, to which d.g is drawn perpendicular. And because the angle d.g.a is right, by the 18th of the first the line a.d is longer than the line d.g, which is impossible, and so no right line may be interposed between it and the circumference.	¶ Cadet ergo extra sitque .a.e. quod si inter ipsam et circumferentiam potest linea recta intercipi sit illa .a.f. ad quam ducatur perpendicularis .d.g. et quia angulus .d.g.a. est rectus. erit per .18. primi linea .a.d. longior linea .d.g. quod est impossibile. quare inter ipsam et circumferentia nulla linea recta intercipietur.
¶ Because of that it is clear that the angle contained by e.a and the circumference, which is said to be the angle of contiguity, is less than any angle contained by two right lines.	¶ Propter quod patet quod angulus contentus. ab .e.a. et circumferentia qui dicitur angulus contingentiae est minor omni angulo a duabus rectis lineis contento.
For if any right-lined angle were to be equal to the angle of contiguity, or less than it, since every such angle is able to be divided by equals following the 9th doctrine of the first, a right line between a.e and the circumference would be able to be interposed, which we have shown to be not possible, by which it is evident the angle contained by the diameter and circumference is to be greatest of all acute right-lined angles because it does not differ from a right outside respecting the angle of contiguity, which we have shown to be smaller than every right-lined angle.	Si enim aliquis rectilineus angulus esset angulo contingentiae aequalis. aut eo minor cum omnis talis possit per aequalia dividia secundum doctrinam .9. primi inter lineam .a.e. et circumferentiam posset linea recta intercipi quod monstravimu esse non posse. per quod patet angulum contentum a diametro et circumferentia omnium acutorum rectilineorum esse maiorem. quia non differt a recto nisi in angulo contingentiae quem monstravimus esse minorem omni rectilineo.
¶ The corollary is evident by the first part.	¶ Correlarium patet per primam partem.
For when the line a.e produced at either end does not cut the circle and touches it at point a, it is contacting it by definition.	Cum enim linea .a.e. in utraque partem eiecta non secet circulum et tangat ipsum in puncto .a. ipsa est contingens per diffinitionem.
¶ From here it should be noted that this argumentation is not sound; this passes from lesser to greater and throughout all the middle, and so through the equal. Nor is this, which affects to find something greater than this and less than the same, and so affects to find the equal, and this is evident as follows:	¶ Ex hoc notandum quod non valet ista argumentatio. hoc transit a minori ad maius. et per omnia media. ergo per aequale. nec ista contingit reperire maius hoc et minus eodem ergo contingit reperire aequale. hoc autem sic patet.
Let the circle be a.b, over the center c, whose diameter is a.c.b, and at its terminus a, suppose the line a.d is drawn orthogonally so that it is contacting the circle by the corollary of this.	Sit circulus .a.b. super centrum .c. cuius diameter .a.c.b. et ducatur ab eius termino .a. linea .a.d. orthogonaliter: eritque contingens circulum per correlarium huius.
And again a circle is drawn, b.e.d over the center a, following the extent of the diameter a.b. And the line a.b is imagined to be moved beyond point a, along the circumference of the arc b.e.d, so that point b shall enumerate all the points of the arc b.e.d until it shall reach the line a.d and fully cover it. And because the angle b.a.d is right, it will not be as to presuppose any acute angle to which the line a.b shall not be made equal along with the diameter a.c.b of the smaller circle, because it passed to the right angle, enumerating the position of all acute angles, whence it is manifest some are to be smaller than the angle of the semicircle contained by half the circumference a.b and the diameter a.c.b, and the right angle is clearly to be greater than those.	Describatur iterum super punctum .a. secundum quantitatem diametri .a.b. circulus .b.e.d. et imaginetur linea .a.b. moveri super punctum .a. per circumferentiam arcus .b.e.d. ita quod punctum .b. numeret omnia puncta arcus .b.e.d. quousque perveniat ad lineam .a.d. et cooperiat ipsam. et quia angulus .b.a.d. est rectus: erit ut non sit sumere aliquem angulum acutum cui aequalem non fecerit linea .a.b. cum diametro .a.c.b. minoris circuli. quia transivit ad angulum rectum dinumerans situm omnium angulorum acutorum quorum manifestum est quosdam esse minores angulo semicirculi contento a semicircumferentia .a.b. et diametro .a.c.b. et angulum rectum manifestum est esse maiorem eodem.
I say that no intermediate in passing from the smaller acutes to the largest right shall be equal to it.	Dico quod nullus in transitu ab acutis minoribus ad recto maiorem intermedius fuit ei aequalis.
But if there were any, suppose that it were produced upon line a.b when point b was at point e of the arc b.e.d, and because the angle e.a.b is equal to the aforementioned angle of the semicircle and the angle of the semicircle is the widest of all acute angles by the final part of this, the angle e.a.b will be the widest of all acute angles. Thus, the angle e.a.d shall be divided as proposed by the 9th of the first, by equals at the line drawn a.f, and by the conception, the angle f.a.b will be wider than the angle e.a.b, thus, will be more wide than the widest, which is impossible.	Si enim fuerit aliquis: sit ut illum fecerit linea .a.b. cum punctus .b. fuit in puncto .e. arcus .b.e.d. quia ergo angulus .e.a.b. est aequalis angulo semicirculi praedicto: angulus autem semicirculi est amplissimus omnium acutorum per ultimam partem huius: erit angulus .e.a.b. amplissimus omnium acutorum: dividatur ergo angulus .e.a.d. sicut proposuit .9. primi per aequalia ducta linea .a.f. eritque per conceptionem angulus .f.a.b. amplior angulo .e.a.b. quare erit aliquid amplius amplissimo quod est impossibile.
¶ Or thus, since the angle e.a.b shall be equal to the angle of the semicircle, just as posited.	¶ Vel sic cum angulus .e.a.b. sit aequalis angulo semicirculi sicut ponitur.
Whereas the angle of the semicircle with the angle of contiguity is equal to one right.	At angulus simicirculi cum angulo contingentiae est aequalis uni recto.
Similarly, the angle e.a.b with the angle e.a.d is also equal to one right, and the angle e.a.d will equal the angle of contiguity. And because the angle of contiguity is most narrow of all acutes, by the 3rd part of this, likewise the angle e.a.d will be equal to the widest of all acute angles. But the angle e.a.f is more narrow to it by the conception, therefore it will be more narrow than the narrowest, which is impossible.	Similiter quoque angulus .e.a.b. cum angulo .e.a.d. est aequalis uni recto: erit angulus .e.a.d. aequalis angulo contingentie: et quia angulus contingentiae est angustissimus omnium acutorum per .3. partem huius: erit similiter angulus .e.a.d. sibi aequalis angustissimus omnium acutorum. set angulus .e.a.f. est eo angustior per conceptionem: erit ergo aliquid angustius angustissimo quod est impossibile.
¶ And so the right-lined angle is not equal to the angle of the semicircle, and that it passes from least to greatest and not through the equal, and likewise because it is to find the lesser and greater, there is clear instance against both preceding arguments.	¶ Non ergo erit angulus rectilineus aequalis angulo semicirculi et quia transitur a minori ad maius et non per aequale. Item quia est reperire minorem eo et maiorem: patet instantia contra utramque argumentationem praedictam.
From which that is to be responded to by an interruption.	Unde per interemptionem ad illud est respondendum.
¶ It could be demonstrated that the angle of contiguity is divisible by a right line, as composed by the figure positioned here on the side.	¶ Posset probari quod angulus contingentiae est divisibilis secundum lineam rectam ut constat per figurationem hic a latere positam.
It is certain that the angle that is caused by the contact of two circles (or of spheres) is the angle of contiguity, and such is divided by the line e.g because that is held by the triangle h.g.k, whose base is h.k, and it is divided by equals at point e and is extended toward g, touching it, and it is argued by the 4th of the first and then by the 26th of this, and the proposition is clear.	Certum est quod angulus qui causatur ex contactu duorum circulorum vel sphaerarum est angulus contingentiae et talis dividatur per lineam .e.g: quia hic habetur triangulus .h.g.k cuius basis h.k dividatur per aequalia in puncto .e. et protrahatur versus .g. contactum et arguitur per .4. primi. deinde per .26. huius et patet propositum.
Proposition 16	Propositio .16.
[English] Latin	English [Latin][¶]
To draw a line contacting a given circle from a given point.	A dato puncto ad datum circulum lineam contingentem ducere.
¶ Suppose the given circle is a.b, whose center is c, and the given point is d, hence I wish to draw a line contacting the circle a.b from point d. I draw the line d.c, cutting the circumference of the circle a.b at point a, over which I draw the circle d.e, following the extent of the line d.c, concentric to the circle a.b. And from point a, I draw line a.e perpendicular to the line d.c, which cuts the circumference of the circle d.e at point e, and I draw the line e.c, cutting the circumference of the circle a.b at point b. Then I draw the line d.b, which will be contacting the circle a.b such that the two sides a.c and c.e of the triangle a.c.e are equal to the two sides b.c and c.d of the triangle b.c.d, and c is the common angle to them both. And by the 4th of the first the angle e.a.c will be equal to the angle d.b.c, but the angle e.a.c is right and so the angle d.b.c is right, and so by a corollary of the preceding the line d.b will be contacting the circle a.b, which is the intent.	¶ Sit circulus datus .a.b. cuius centrum .c. punctusque datus .d. volo ergo a puncto .d. ducere lineam contingentem circulum .a.b. produco lineam .d.c. secantem circumferentiam circuli .a.b. in puncto .a. super quam describo circulum .d.e. secundum quantitatem lineae .d.c. concentricum circulo .a.b. et a puncto .a. produco lineam .a.e. perpendicularem ad lineam .d.c. quae secet circumferentiam circuli .d.e. in puncto .e. et produco lineam .e.c. secantem circumferentiam circuli .a.b. in puncto .b. deinde produco lineam .d.b. quae erit contingens circulum .a.b. quia enim duo latera .a.c. et .c.e. trianguli .a.c.e. sunt aequalia duobus lateribus .b.c. et .c.d. trianguli .b.c.d. et angulus .c. est communis utrique erit per .4. primi angulus .e.a.c. aequalis angulo .d.b.c. angulus autem .e.a.c. est rectus quare angulus .d.b.c. est rectus: per correlarium ergo praecedentis erit linea .d.b. contingens circulum .a.b. quod est propositum.
Proposition 17	Propositio .17.
[English] Latin	English [Latin][¶]
If a right line contacts a circle and from the point of contact a right line is drawn to the center, it is necessary for it to be perpendicular to the contacting line.	Si circulum linea recta contingat a contactu vero ad centrum linea recta ducatur. necesse est eam super lineam contingentem esse perpendicularem.
¶ Suppose the line a.b is contacting the circle c.e, whose center is d, at point c, which is joined to the center by the line c.d. I say this is to be perpendicular to the contacting line.	¶ Sit linea .a.b. contingens circulum .c.e. cuius centrum sit .d. in puncto .c. qui iungatur cum centro per lineam .c.d. dico hanc esse perpendicularem super lineam contingentem.
For if it is not perpendicular to that, then suppose d.f is perpendicular to the same, which shall cut the circumference of the circle at point e. And both of the angles that are against f will be right and so by the 18th of the first, the line c.d is greater than the line d.f, which is impossible.	Si enim non est perpendicularis ad ipsam. sit ergo .d.f. perpendicularis ad eandem quae secet circumferentiam circuli in puncto .e. eritque uterque angulorum qui sunt. ad .f. rectus igitur per .18. primi linea .c.d. est maior linea .d.f. quod est impossibile:
d.c is to be thus perpendicular to a.b, which is the intent.	Constat itaque .d.c. esse perpendicularem super .a.b. quod est propositum.
Proposition 18	Propositio .18.
[English] Latin	English [Latin][¶]
If a right line contacts a circle and from the point of contact somesuch line is drawn orthogonally into the circle, it is necessary to likewise be within the center.	Si circulum linea recta contingat: et a contactu in circulum linea quaedam orthogonaliter ducatur. in eadem centrum esse necesse est.
¶ Suppose first that the line a.b is contacting the circle c.e at point c, and from the point of contact a line is drawn inside the circle c.e, perpendicular to the line a.b. I say that the center of the circle is along the line c.e, and is a converse of the former.	¶ Sit ut prius linea .a.b. contingens circulum .c.e. in puncto .c. et. a contactu ducatur intra circulum .c.e. linea perpendicularis ad lineam .a.b. dico quod centrum circuli est in linea .c.e. et est conversa prioris.
But if the center were not upon the line c.e, then let it happen wheresoever elsewhere and suppose that is at d. And the line d.c is drawn, and by the preceding d.c is perpendicular to the line a.b, which is impossible since e.c is stated perpendicular to that, thus the proposition is clear.	Si enim non fuerit centrum in linea .c.e. sit alibi ubicunque contingat. sitque .d. et producatur linea .d.c. eritque .d.c. per praemissam perpendicularis ad lineam .a.b. quod est impossibile cum .e.c. posita sit perpendicularis ad ipsam: quare patet propositum.
Proposition 19	Propositio .19.
[English] Latin	English [Latin][¶]
If an angle within a circle rests upon the center and another angle resting upon the circumference has the same base, the inferior will be twice the superior.	Si intra circulum angulus supra centrum consistat: alius vero angulus supra circumferentiam consistens eandem basim habeat inferior superiori duplus erit.
¶ Suppose that within the circle a.b.c, whose center is d, the angle a.d.c is upon the center and the angle a.b.c is upon the circumference, and suppose the base of each of the angles is the same, and that shall be the arc a.c. I say the angle a.d.c is to be twice the angle a.b.c.	¶ Sit ut in circulo .a.b.c. cuius centrum .d. fiat angulus .a.d.c. super centrum et angulus .a.b.c. super circumferentiam. sitque utriusque anguli eadem basis qui sit arcus .a.c. dico angulus .a.d.c. duplum esse ad angulum .a.b.c.
¶ That is demonstrated as follows. For either the two lines a.b and b.c include the two lines a.d and d.c, or one of the first is made as one line with one of the remaining, or one of the former cuts one of the latter.	¶ Quod sic probatur. Aut enim duae lineae .a.b. et .b.c. includunt duas lineas .a.d. et .d.c. aut altera earum fit linea una cum altera reliquarum. aut etiam altera primarum secat alteram postremarum.
¶ Therefore, first suppose that it includes them, just as appears in the first figure, and the line b.d.e is drawn, and by the 32nd of the first the angle a.d.e from without is equal to the two within, which are the angles b.a.d and a.b.d. And that they are equal by the 5th of the same, the angle a.d.e will be twice the angle a.b.d and similarly the angle e.d.c will be twice the angle d.b.c, hence the whole angle a.d.c is twice the whole angle a.b.c, which is the intent.	¶ Sit ergo primo ut includant eos ut in prima figuratione apparet. et producatur linea .b.d.e. eritque per .32. primi: angulus .a.d.e. extrinsecus aequalis duobus intrinsecis qui sunt .b.a.d. et .a.b.d. anguli et quia ipsi sunt aequales per .5. eiusdem erit angulus .a.d.e. duplus ad angulum .a.b.d. similiter quoque erit angulus .e.d.c. duplus ad angulum .d.b.c. quare totus angulus .a.d.c. duplus est ad totum angulum .a.b.c. quod est propositum.
¶ And if one of the two lines a.b and b.c were one line with another of the two lines, which are a.d and d.c, as appears in the second figure, the proposition is clear by the same means as before and in a similar way.	¶ Quod si altera duarum linearum .a.b. et .b.c. fiat linea una cum altera duarum linearum quae sunt .a.d. et .d.c. ut in secunda figuratione apparet. per easdem per quas prius: et simili modo liquet propositum.
¶ But if one of the first two cuts another of the two remaining, as appears in the 3rd figure when the line a.b cuts the line d.c, the line b.d.e is produced. And by the same means we previously assumed, and in a similar way, the angle e.d.a will be double the angle d.b.a and the whole angle e.d.c will be double the whole angle d.b.c, hence the angle a.d.c is twice the angle a.b.c, which is the intent.	¶ Quod si altera duarum primarum secet alteram duarum postremarum. ut in .3. figuratione apparet. ubi linea .a.b. secat lineam .d.c. producatur linea .b.d.e. eritque per easdem quas prius assumpsimus et simili modo angulus .e.d.a. duplus ad angulum .d.b.a. et totus angulus .e.d.c. duplus ad totum angulum .d.b.c. quare angulus .d.b.c.[sic] duplus est ad angulum .a.b.c. quod est propositum.
Proposition 20	Propositio .20.
[English] Latin	English [Latin][¶]
If angles rest upon the arc within one portion of a circle, it is necessary for all the angles to be equal.	Si in una circuli portione anguli super arcum consistant angulos quoslibet esse aequales necesse est.
¶ Suppose that within the portion a.d.b of the circle a.d.b, whose center is f, the angles all rest upon the arc a.d.b, which are c, d, and e. I say they are to be equal. For the chord a.b is drawn and from its extremities the lines a.f and b.f are drawn to the center, and through the preceding the angle f will be resting upon the center and will be twice each of them: wherefore they are equal, which is the intent.	¶ Sit ut in portione .a.d.b. circuli .a.d.b. cuius centrum .f. consistant quoslibet anguli super arcum .a.d.b. qui sunt .c.d.e. dico eos esse aequales. protrahatur enim corda .a.b. et ab eius extremitatibus: ducantur in centrum lineae .a.f. et .b.f. eritque per praemissam angulus .f. consistens super centrum ad unumquemque eorum. duplus: quare ipsi sunt aequales: quod est propositum.
Proposition 21	Propositio .21.
[English] Latin	English [Latin][¶]
If a quadrilateral is drawn within a circle, it is necessary for any two of its angles taken from opposite sides to be equal to two rights.	Si intra circulum quadrilaterum describatur. quoslibet eius duos angulos ex adverso collocatos duobus rectis angulis equos esse necesse est.
¶ Suppose the quadrilateral a.b.c.d is drawn within the circle a.b.c.d. I say each two of its angles assembled from opposite sides are equal to two rights, for within the quadrilateral the diameters a.c and b.d are drawn, and by the preceding the angle c.b.d equals the angle c.a.d and the angle a.b.d equals a.c.d, whence the whole a.b.c will be equal to the two angles that are a.c.d and c.a.d, and because those with the angle a.d.c are equal to two rights by the 32nd of the first, the two angles, the total b and the total d, will be equal to two rights, which is the intent.	¶ Sit quadrilaterum .a.b.c.d. inscriptum circulo .a.b.c.d. dico quosque duos eius angulos ex adverso collocatos esse aequales duobus rectis. protrahantur enim in quadrilatero diamertri .a.c.b.d. eritque per praemissam angulus .c.b.d. aequalis angulo .c.a.d. et angulus .a.b.d. angulo .a.c.d. quare totus .a.b.d. [sic] erit aequalis duobus angulis qui sunt .a.c.d. et .c.a.d. et quia ipsi cum angulo .a.d.c. sunt aequales duobus rectis. per .32. primi: erunt duo anguli .b. totalis. et .d. totalis aequalis duobus rectis quod est propositum.
It will likewise be similarly proved the total angles a and c are equal to two rights.	Similiter quoque probabitur angulos .a. et .c. totales esse aequales duobus rectis.
Proposition 22	Propositio .22.
[English] Latin	English [Latin][¶]
It is impossible for two similar, unequal portions of a circle on one assigned right line to fall from the same side.	Duas circuli similes portiones inaequales. super unam rectam lineam assignatam. ex eadem parte cadere impossibile est.
¶ Suppose the assigned right line is a.b, over which the portion of the circle a.c.b is made. I say that there shall be no other portion which is similar to this on the same line, from the same side, and it shall either be greater or lesser.	¶ Sit linea recta assignata .a.b. super quam fiat portio circuli .a.c.b. dico quod super eandem lineam ex parte eadem non fiet alia portio quae sit similis huic. et ea maior aut minor.
¶ But if this were possible, then suppose the portion a.d.b, which is greater, to be nonetheless similar to it, and so the angle a.c.b shall be made in the smaller portion and the angle a.d.b in the larger, so the lines a.d and d.b will enclose the lines a.c and c.b as is apparent in the first figure.	¶ Quod si fuerit hoc possibile fiat ergo portio .a.d.b. maior ea quae tamen sit similis ei. fiat ergo angulus .a.c.b. in portione minori. et angulus .a.d.b. in maiori. erit ergo ut lineae .a.d. et d.b. includant lineas .a.c. et .c.b. ut patet in figuratione prima.
Or so one of the former may be made to be the same with one of the latter as in the second, or so one cuts the other as in the third.	Aut ut altera primarum fiat eadem cum altera postremarum. ut in secundam. aut ut altera secet alteram ut in tertia.
¶ So if it were in the first way, by the 21st of the first angle c will be greater than angle d, thus they are not similar portions by definition.	¶ Quod si fuerit primo modo erit per .21. primi: angulus .c. maior angulo .d. non ergo sunt portiones similes per diffinitionem.
And if in the second way, angle c will still be greater than angle d by the 16th of the same, and so it does not follow then that the portions will be similar.	Quod si secundo modo erit adhuc angulus .c. maior angulo .d. per .16. eiusdem. nec sic igitur erunt portiones similes.
But if in the third way, suppose that the line b.d cuts the line a.c and cuts the circumference of the small portion at point e, and say the line e.a is drawn. Then by that same 16th of the first, the angle a.e.b in portion a.c.b will be greater than the angle d (e is obviously equal to c by the 20th of this, whence c is greater than d) and so is in no way similar.	Si autem tertio modo sit ut linea .a.d. [sic] secet lineam .c.b. [sic] et secet circumferentiam portionis minoris in puncto .e. et ducatur linea .e.b. [sic] eritque per eandem .16. primi: angulus .a.e.b. consistens in portione .a.c.b. maior angulo .d. scilicet .e. est aequalis .c. per .20. huius quare .c. est maior .d. quare nullo modo similes.
¶ And in the same way you will prove that there shall not be a portion similar to the portion a.c.b on the line a.b, smaller than it, by placing c in place of d and d in place of c on the previous figures, for through the preceding it is evident by the 21st and the 16th of the first, as well as by the foregoing way in which the angle d of every configuration is greater than angle c, wherefore the portions will not be similar.	¶ Simili quoque modo probabis quod super linea .a.b. non fiet portio similis portioni .a.c.b. minor .e.a. [sic] posito .c. in loco .d. et .d. in loco .c. in figurationibus praedictis. erit enim per praemissas scilicet per .21. et .16. primi: et praemisso modo angulus .d. omnium figurationum maior angulo .c. quare portiones non erunt similes.
¶ And note that while it's proposed to be not possible for similar, unequal portions to be made on one line from the same side, it is yet true that neither may it be proved from the opposite side; the smaller, which is from one side, superimposed over the larger, which is from the other. For through common knowledge it will be necessary that the smaller is to be exceeded by the larger, and so they are not similar by this, the 22nd.	¶ Et nota quod licet proponatur super lineam unam non posse fieri portiones similes inaequales ex eadem parte. verum est tamen quod nec ex diversis quod licet probare minori quae est ex una parte superposita maiori quae est ex alia. necesse enim. erit per communem scientiam ipsam a maiori excedi. non ergo sunt similes per hanc .22.
Proposition 23	Propositio .23.
[English] Latin	English [Latin][¶]
If similar portions of circles shall be on equal lines, it is necessary those portions are to be equal.	Si circulorum similes portiones super lineas aequas fuerint. ipsas portiones aequales esse necesse est.
¶ Suppose the two lines a.b and c.d are equal, over which are two portions of the circles a.e.b and c.f.d, which are similar. I say that they are equal, for if they are not equal and one of them were superimposed over the other, then the greater shall exceed the lesser. But the line a.b does not exceed the line c.d, nor is it exceeded by it, beings as they are equal, wherefore it falls contrary to the premise, which is impossible, because a.b and c.d shall be one line.	¶ Sint duae lineae .a.b. et .c.d. aequales super quas sint duae portiones circulorum .a.e.b.c.f.d. quae sunt similes. dico quod ipsae sunt aequales. si enim non sunt aequales altera earum superposita alteri excedet maior minorem. sed linea .a.b. non excedet lineam .c.d. nec excedetur ab ea: cum sint aequales. quare accidit contrarium praemissae quod est impossibile. erunt enim a.b. et .c.d. linea una.
Proposition 24	Propositio .24.
[English] Latin	English [Latin][¶]
To complete the circle of a given semicircle, or of a portion either less or greater than a semicircle.	Dati semicirculi. sive semicirculo maioris minorisve portionis, circulum perficere.
¶ It is intended by way of this conclusion to complete a circle from any given arc or from any given portion of a circle.	¶ Intentum per hanc conclusionem est ex omni arcu dato sive ex omni circuli portione data perficere circulum.
And so suppose a.b to be any arc from which I wish to complete a circle. I will draw in it two lines, howsoever it may come to pass, which are a.c and b.d, and will divide those by equals, that is, a.c at point e and b.d at point f. And I will draw e.g perpendicular to a.c and f.h perpendicular to b.d, which shall cut one another at point k, and by a corollary of the first of this the center of the circle is upon both of the lines e.g and f.h, whence the center is point k.	Sit ergo .a.b. quilibet arcus ex quo volo perficere circulum protraham in eo duas lineas qualitercunque contingat quae sint .a.c. et .b.d. quas dividam per aequalia a.c. quidam in puncto .e. et .b.d. in puncto .f. et protraham .e.g. perpendicularem ad .a.c. et .f.h. perpendicularem ad .b.d. quae secent se in puncto .k. eritque per correlarium primae huius centrum circuli in utraque linearum .e.g. et .f.h. quare centrum est punctum .k.
And if e.g does not cut f.h but they are one line, just as it would be if the two lines a.c and b.d were equidistant, then it will be applied to the circumference of the given arc from both sides and so it is divided by half at point k and there the center of the circle shall be by the same corollary.	Si autem .e.g. non secet .f.h. sed sint linea una. quemadmodum erit si duae lineae .a.c. et .b.d. sint aequidistantes tunc ipsa applicabitur circumferentiae dati arcus. ex utraque parte ipsa igitur divisa per medium in puncto .k. erit ibi centrum circuli per idem correlarium.
However, e.g and f.h will not be equidistant, for when the center of the circle is within each by the stated corollary, there would be two centers of the same circle.	Aequidistantes autem non erunt .e.g. et .f.h. quia cum in utraque sit centrum circuli per dictum correlarium. essent eiusdem circuli duo centra.
¶ Nevertheless, because the author seems to vary this conclusion according to the different types of arcs by enumerating the types of all the portions, we may demonstrate separately by type how a circle is completed from any given portion.	¶ Quia tamen auctor videtur hanc conclusionem variare secundum diversas species arcuum omnium portionum enumerando species: demonstrabimus divisim per species qualiter ex omni portione data circulus perficiatur.
And so first suppose the given portion a.b is a semicircle, and by the definition of a semicircle the line a.b is the diameter. Thus, it is divided by half at point c and c will be the center of the circle.	Sit ergo primum .a.b. portio data semicirculus. eritque per diffinitionem semicirculi. linea .a.b. diametri ea igitur divisa per medium in puncto .c. erit .c. centrum circuli.
¶ Again suppose the portion a.c.b is greater than a semicircle, whose chord is a.b, which I divide by equals at point d, and from that I draw d.c perpendicular to it, which will travel through the center by a corollary of the first of this. And I extend the line a.c, and that the line a.b is smaller than the diameter, because the portion a.c.b is greater than a semicircle a.d will be smaller than half the diameter. But d.c is greater than half the diameter and so d.c is greater than a.d, therefore by the 19th of the first the angle c.a.d is greater than the angle a.c.d. Thus, let it be by the 23rd of the first that the angle c.a.e equals the angle a.c.d. The line a.e, having been extended, cuts the line c.d at point e, and by the sixth of the first the line a.e equals the line e.c. And so were the line e.b produced then by the 4th of the first the line e.b will equal the line a.e, whence the three lines, e.a, e.b, and e.c, are equal, and so by the 9th of this e is the center of the circle.	¶ Sit rursus portio .a.c.b. semicirculo maior cuius corda sit .a.b. quam divido per aequalia in puncto .d. a quo duco .d.c. perpendicularem ad ipsam quae transibit per centrum per correlarium. primae huius: et protraho lineam .a.c. et quia linea .a.b. est minor diametro cum sit .a.c.b. portio maior semicurculo: erit .a.d: minor semidiametro. sed .d.c. est maior semidiametro. ergo .d.c. est maior quam .a.d. ergo per .19. primi: angulus .c.a.d. est maior angulo .a.c.d. fiat itaque per .23. primi: angulus .c.a.e. aequalis angulo .a.c.d. producta linea .a.e. quae secet lineam .c.d. in puncto .e. eritque per sextam primi. linea .a.e. aequalis lineae .e.c. producatur igitur linea .e.b. eritque per .4. primi linea .e.b. aequalis lineae .a.e. quare tres lineae .e.a.e.b.e.c. sunt aequales ergo per .9. huius .e. est centrum circuli.
¶ Once more, suppose a.c.b is a portion less than a semicircle, whose chord is a.b, which I divide by equals at point d, from which I draw the line c.d.e perpendicular to the line a.b, and that cuts the circumference at point c. And it is clear that this is to pass through the center by a corollary of the first of this, and again, I draw the line a.c and the angle a.c.d will be greater than the angle c.a.d. If it is equal the portion a.c.b will be a semicircle, and if smaller it will be greater than a semicircle, however it is supposed that it should be smaller, therefore I draw the line a.e, which with the line a.c makes an angle equal to angle c, and it cuts the line c.f at point e and it is clear that point e falls without the given portion and I draw the line e.b. And that the whole angle of a is equal to the angle c, by the 6th of the first the line e.a will equal the line e.c, and that by the fourth of the first the line e.b equals the line e.a, by the 9th of this point e will be the center of the circle, whence the proposition is evident according to all the types of portions of a circle.	¶ Sit iterum .a.c.b. portio minor semicirculo. cuius corda sit a.b. quam divido per aequalia in puncto .d. a quo produco lineam .c.d.e. perpendicularem ad lineam .a.b. quae secet circumferentiam in puncto .c. hanc manifestum est transire per centrum per correlarium primae huius. produco iterum lineam .a.c: eritque angulus .a.c.d. maior angulo .c.a.d. si est aequalis erit portio .a.c.b. semicirculus. et si minor erit maior semicirculo: positum est autem quod sit minor. produco igitur lineam .a.e. quae cum linea .a.c. faciat angulum aequalem angulo .c. et secet lineam .c.f. in puncto .e. et manifestum est quod punctum .e. cadat extra datam portionem. et produco lineam .e.b. et quia angulus .a. totalis est aequalis angulo .c. erit per .6. primi linea .e.a. aequalis lineae .e.c. et quia per quartam primi linea .e.b. est aequalis lineae .e.a. erit per .9. huius punctum .e. centrum circuli quare patet propositum secundum omnes species portionum circuli.
Proposition 25	Propositio .25.
[English] Latin	English [Latin][¶]
If equal angles stand in equal circles, either on the centers or on the circumferences, it is necessary for them to fall upon equal arcs.	Si in aequis circulis seu super centra. seu super circumferentias aequales anguli consistant. super aequos arcus eos cadere necesse est.
¶ Suppose there are two equal circles, a.b.c, whose center is d, and e.f.g, whose center is h, and over the centers of them are made the two angles, a.d.c and e.h.g, which are assumed equal. I say the two arcs, a.b.c and e.f.g, are to be equal. The two lines a.c and e.g shall be drawn and two angles are made on the circumference of them, resting over the predicated arcs, which are the angle a.b.c and the angle e.f.g. And because the circles are equal, by the definition of equal circles their half diameters will be equal. and because the two angles d and h are equal, by the 4th of the first the line a.c will equal the line e.g and by the 19th of this, angle b will equal angle f because angle d is equal to angle h, and so by the definition of similar portions the two portions a.b.c and e.f.g are equal, and because they are over the equal lines a.c and e.g, they are then equal by the 23rd of this, whence the arcs a.b.c and e.f.g are equal.	¶ Sint duo circuli aequales .a.b.c. cuius centrum .d. et .e.f.g. cuius centrum .h. et fiant supra centra eorum duo anguli .a.d.c. et .e.h.g. qui ponantur aequales. dico duas arcus .a.b.c. et .e.f.g. esse aequales. protrahantur duae lineae .a.c. et .e.g. et fiant duo anguli in circumferentiis ipsorum consistentes supra praedictos arcus qui sint angulus .a.b.c. et angulus .e.f.g. quia ergo circuli sunt aequales. erunt per diffinitionem aequalium circulorum semidiametri aequales: et quia duo anguli .d. et .h. sunt aequales erit per .4. primi: linea .a.c. aequales lineae .e.g. et per .19. huius erit angulus .b. aequalis angulo .f. cum .d. angulus sit aequalis angulo .h. ergo per diffinitionem similium portionum duae portiones .a.b.c. et .e.f.g. sunt similies: et quia ipse sunt super lineas: a.c. et .e.g. aequales ipse erunt aequales per .23. huius: quare arcus .a.b.c. et .e.f.g. sunt aequales.
And if the angles b and f are on the circumference, they are assumed equal by the definition of similar portions, and the angle d and h are equal by the 19th of this. And because the circles are equal by position, by the 4th of the first the two lines a.c and e.g are equal, and so as before, the portions are equal by the 23rd of this because they are similar and are over equal lines, the arcs are therefore equal, which is the intent.	Quod si anguli .b. et .f. qui sunt in curcumferentia ponantur aequales erunt per diffinitionem portiones siiles et anguli .d. et .h. aequales per .19. huius: et quia circuli sunt aequales per positionem erunt per .4. primi: duae lineae .a.c. et .e.g. aequales quare ut prius portiones aequales per .23. huius cum sint similes et super aequales lineas. igitur et arcus aequales: quod est propositum.
Proposition 26	Propositio .26.
[English] Latin	English [Latin][¶]
If equal arcs are assumed on equal circles, it is necessary for the angles fashioned beneath them, which are placed either on their centers or on their circumferences, to be equal.	Si in aequis circulis aequi sumantur arcus. infra illos formatos angulos. qui supra centra eorum seu supra circumferentias constituantur aequos esse necesse est.
¶ First suppose that there are two equal circles, a.b.c whose center is d and e.f.g whose center is h, and there are two equal arcs, a.b.c and e.f.g, and over those arcs two angles are made at the center, which are d and h, by drawing a.d, c.d, e.h, and g.h.	¶ Sint ut prius duo circuli .a.b.c. cuius centrum .d. et .e.f.g. cuius centrum .h. sintque duo arcus .a.b.c. et e.f.g. aequales fiantque super ipsos arcus duo anguli in centro qui sint .d. et .h. ductis .a.d.c.d.e.h.g.h.
Likewise, over the same arc two other angles are made at the circumferences, which are b and f, by drawing the lines a.b, c.b, e.f, and g.f. I say the two angles d and h are mutually equal.	Itemque super eosdem arcus fiant duo alii anguli in circumferentia qui sint .b. et .f. ductis lineis .a.b.c.b.e.f. et .g.f. dico duos angulos .d. et .h. ad invicem esse aequales.
And also, the two angles b and f are mutually equal, and this is the converse of the preceding. But if angles d and h are not mutually equal then suppose h is greater, from which the angle k.h.g is abscinded, which shall be equal to angle d. And through the preceding the arc k.e.f.g equals the arc a.b.c. But the two arcs a.b.c and e.f.g are assumed equal, therefore, it will happen that a part is to be equal to the whole, which is impossible, whence the whole angles d and h are equal.	Itemque duos .b. et .f. adinvicem esse aequales et est haec conversa prioris. si enim non sunt .d. et .h. anguli ad invicem aequales: sit ergo .h. maior a quo abscindatur angulus .k.h.g. qui sit aequalis angulo .d. eritque per praemissam arcus .k.e.f.g. aequalis arcui .a.b.c. sed duo arcus .a.b.c. et .e.f.g. positi sunt aequales: accidet ergo partem esse aequalem toti: quod est impossibile: quare anguli .d. et .h. totales sunt aequales.
¶ In a similar way you will also demonstrate angles b and f to be equal, or prove that if the greater angles d and h are equal, it follows for b and f to be equal by the 19th of this and vice versa.	¶ Simili quoque modo probabis angulos .b. et .f. esse aequales. vel si maius probato quod anguli .d. et .h. sint aequales. sequitur .b. et .f. esse aequales per .19. huius et econverso.
Proposition 27	Propositio .27.
[English] Latin	English [Latin][¶]
If equal lines resect arcs within equal circles, the arcs too shall be equal. And if the lines are not equal then the arcs too shall not be equal, and it is necessary for a greater arc to be abscinded by a greater line and a lesser arc by a lesser line.	Si in circulis aequalibus aequae lineae arcus resecent. arcus quoque aequos esse. si autem lineae inaequales fuerint arcus quoque inaequales. et a maiore linea maiorem arcum: a minore vero minorem abscindi necessarium est.
¶ Suppose there are two equal circles, a.b.c whose center is d and e.f.g whose center is h, and the cord a.c equals the cord e.g. I say the two arcs, a.b.c and e.f.g, which the aforesaid cords resect from aforesaid circles, are to be equal.	¶ Sint duo circuli aequales .a.b.c. cuius centrum .d. et .e.f.g. cuius centrum .h. sitque corda .a.c. aequalis corde .e.g. dico duos arcus .a.b.c. et .e.f.g. quos praedictae cordae ex praedictis circulis resecant esse aequales.
But if the cord e.g were supposed greater than the cord a.c, I say the arc e.f.g is to be greater than the arc a.b.c.	Quod si corda .e.g. ponatur maior corda .a.c. dico arcum .e.f.g. esse maiorem arcu .a.b.c.
The first is in fact demonstrated thus: lines are drawn from the centers to the extremities of the chords, which are d.a, d.c, h.e, and h.g, and because the circles are posited to be more or less equal, their half diameters will be equal. And because the line a.c is posited equal to the line e.g, by the 8th of the first the angle d will equal the whole angle h, whence by the 25th of this the arc a.b.c will equal the arc e.f.g, and so the first is evident. The second is thus: suppose e.g greater than a.c, and so by the 25th of the first the angle h will be greater than the angle d and so the angle f.h.g shall equal the angle d, and by the 25th of this the arc f.g is equal to the arc a.b.c, whence the arc e.f.g is greater than the arc a.b.c, and that is the second intention.	Primum quidem sic probatur ducantur a centris lineae ad extremitates cordarum quae sint .d.a.d.c.h.e.h.g. et quia circuli positi sunt fere aequales. erunt hea semidiametri aequales. et quia linea .a.c. posita est aequalis lineae .e.g. erit per .8. primi: angulus .d. aequalis angulo .h. totali: quare per .25. huius erit arcus .a.b.c. aequalis arcui .e.f.g. sicque patet primum. secundum sic. sit .e.g. maior .a.c. eritque per .25. primi angulus .h. maior angulo .d. fiat ergo angulus .f.h.g. aequalis angulo .d. eritque per .25. huius arcus .f.g. aequalis arcui .a.b.c. quare arcus .e.f.g. est maior arcu .a.b.c. quod est secundum. propositum.
Proposition 28	Propositio .28.
[English] Latin	English [Latin][¶]
It is necessary for equal arcs of equal circles to have equal chords.	Circulorum aequalium aequos arcus. aequas cordas habere necesse est.
¶ Suppose the two equal circles, a.b.c, whose center is d, and e.f.g, whose center is h, and suppose the arc a.b.c equals the arc e.f.g. I say that the chord a.c equals the chord e.g, and this is the converse of the first part of the preceding.	¶ Sint duo circuli aequales .a.b.c. cuius centrum .d. et .e.f.g. cuius centrum .h. sitque arcus .a.b.c. aequalis arcui .e.f.g. dico quod corda .a.c. est aequalis cordae .e.g. et est haec conversa primae partis praemissae.
¶ The lines d.a, d.c, h.e, and h.g are drawn, and by the 26th of this the angles d and h will be equal, whence by the fourth of the first a.c will equal e.g, which is the intent.	¶ Ducantur lineae .d.a.d.c.h.e.h.g. eruntque per .26. huius anguli .d. et .h. aequales: quare per quartam primi: erit .a.c. aequalis .e.g. quod est propositum.
But whatsoever phenomena already proved with respect to separate and equal circles, understand there to be a great many more truths about them.	Quaecunque autem probatae sunt passiones de diversis circulis aequalibus intellige multo fortius veras esse de eodem.
Proposition 29	Propositio .29.
[English] Latin	English [Latin][¶]
To divide a given arc by equals.	Datum arcum per aequalia dividere.
¶ Suppose the given arc is a.b.c, which is subtended by the chord a.c, that is divided by equals at point d, from which a line is drawn perpendicular to it, and that shall cut the circumference of the given arc at point b. I say it is to divide the given arc by equals, for the lines b.a and b.c are drawn, which shall be equal by the 4th of the first, whence by the first part of the 27th of this the arc a.b will equal the arc b.c, which is the intent.	¶ Sit datus arcus .a.b.c. cui subtendatur corda .a.c. quae dividatur per aequalia in puncto .d. a quo ducatur perpendicularis ad ipsam quae sit .d.b. secans circumferentiam dati arcus in puncto .b. quem dico dividere datum arcum per aequalia. ducantur enim lineae .b.a.b.c. quae erunt aequales per .4. primi quare per primam partem .27. huius arcus .a.b. erit aequalis arcui .b.c. quod est propositum.
Proposition 30	Propositio .30.
[English] Latin	English [Latin][¶]
If a right-lined angle in a semicircle rests upon the arc, it is right. And if it's in a portion less than a semicircle, it's greater than right. And if it's in a portion greater than a semicircle, it's less than right. Furthermore, the angle of any portion greater than a semicircle will be greater than right and by necessity, that of a lesser portion will be less than right.	Si rectilineus angulus in semicirculo supra arcum consistat. rectus est. Si vero in portione semicirculo minore recto maior. Si autem in portione semicirculo maiore recto minor. Itemque omnis portionis angulus semicirculo maioris recto maior. minoris vero recto minor de necessitate erit.
¶ Suppose that in the circle a.b.c, whose center is d and diameter a.d.c, is the semicircle a.b.c, in of which the angle a.b.c shall be upon the circumference of the semicircle by drawing the lines a.b and b.c. I say that angle is right. The line b.d is drawn from that angle to the center and by the fifth of the first, the angle a.b.d equals angle a and the angle d.b.c equals angle c, and because the angle c.d.b is equal to the two angles d.b.a and a by the 32st of the first, it will be twice the angle d.b.a. By the same rationale the angle a.d.b will be twice the angle d.b.c, therefore, the two angles c.d.b and a.d.b are twice the entire angle a.b.c. But they themselves are equal to two rights by the 13th of the first, and so the entire angle a.b.c will be half of two rights, wherefore is right, which is the first intention.	¶ Sit ut in circulo .a.b.c. cuius centrum .d. et diameter .a.d.c. semicirculus .a.b.c. in cuius semicirculi circumferentia fiat angulus .a.b.c. ductis lineis .a.b. et .b.c. dico illum angulum esse rectum. protrahatur ab ipso angulo in centrum linea .b.d. eritque per quintam primi: angulus .a.b.d. aequalis angulo .a. et angulus .d.b.c. aequalis angulo .c. et quia angulus .c.d.b. est aequalis duobus angulis .d.b.a. et .a. per .32. primi: ipse erit duplus ad angulum .d.b.a. eadem ratione angulus .a.d.b. duplus erit ad angulum .d.b.c. ergo duo anguli .c.d.b. et .a.d.b. dupli sunt ad totalem angulum .a.b.c. sed ipsi sunt aequales duobus rectis. per .13. primi: erit igitur angulus .a.b.c. totalis medietas duorum rectorum: quare rectus quod est primum propositum.
¶ The same differently: b.c is extended up until e and by the 32nd of the first the angle a.b.e equals the two angles a and c, and because angle a is equal to the angle a.b.d and angle c to the angle c.b.d, the angle a.b.e will equal the entire angle a.b.c, and so either of them is right by definition.	¶ Idem aliter protrahatur .b.c. usque ad .e. eritque per .32. primi: angulus .a.b.e aequalis duobus angulis .a. et .c. et quia angulus .a. est aequalis angulo .a.b.d. et angulus .c. angulo .c.b.d. erit angulus .a.b.e. aequalis totali angulo .a.b.c. ergo uterque eorum est rectus per diffinitionem.
¶ The second is evident as follows: in the circle a.b.c, whose center is d, suppose the portion a.b.c, whose chord is a.c, to be greater than a semicircle, and the angle a.b.c shall be upon its circumference by drawing the lines b.a and b.c. I say that angle is less than right, for the line of the diameter a.d.e and the line e.b are drawn, and by the first part of this the whole of b is right, whereby the angle a.b.c will be less than right by the common conception, since it is a part of it, and so the second is clear.	¶ Secundum sic patet: sit in circulo .a.b.c. cuius centrum .d. portio .a.b.c. cuius corda .a.c. maior semicirculo: et fiat super eius circumferentiam angulus .a.b.c. ductis lineis .b.a. et .b.c. dico illum angulum esse minorem recto. ducantur enim diametri .a.d.e. et linea .e.b. eritque per primam partem huius .b. totalis rectus. quare angulus .a.b.c. erit minor recto per communem scientiam cum sit pars eius: sicque patet secundum.
¶ The third, thus:	¶ Tertium sic.
Again suppose in the circle a.b.c, whose center is d, the portion a.b.c that is less than a semicircle, whose chord is a.c, and the angle a.b.c shall be upon its circumference by drawing the lines b.a and b.c. I say this angle is greater than right, for the line of the diameter a.d.e and the line b.e are drawn and by the first part of this the angle a.b.e is right, whereby the angle a.b.c will be greater than right, which is the third intention.	Sit rursus in circulo a.b.c. cuius centrum .d. portio .a.b.c. cuius corda .a.c. quae sit semicirculo minor: et fiat super eius circumferentiam angulus .a.b.c. ductis lineis .b.a. et .b.c. dico hunc angulum esse maiorem recto. producantur enim diametri .a.d.e. et linea .b.e. eritque per primam partem huius angulus .a.b.e. rectus. quare angulus .a.b.c. erit maior recto quod est tertiam propositum.
¶ The forth and fifth thusly:	¶ Quartum et quintum sic.
In the circle a.b.c.d, whose center is e, suppose the portion a.b.c, whose chord is a.c, to be greater than a semicircle, and the portion a.d.c, whose chord is likewise a.c, to be less than a semicircle. I say the angle contained by the arc b.a and the chord a.c is greater than right, and the angle contained by the arc d.a and the chord a.c is less than right. The line of the diameter is drawn and the line b.a extended up until f, and by the first part of this the angle b.a.c is right.	Sint in circulo .a.b.c.d. cuius centrum .e. portio .a.b.c. cuius corda .a.c. maior semicirculo et portio .a.d.c. cuius eadem corda .a.c. minor semicirculo dico angulum contentum ab arcu .b.a. et corda .a.c. esse maiorem recto et angulum contentum ab arcu .d.a. et corda .a.c. esse minorem recto. producatur diametri .c.e.b. et linea .b.a. usque ad .f. eritque per primam partem huius angulus .b.a.c. rectus.
And because the right angle is a part of the first and the second is a part of the right, both are clearly evident, whence the whole conclusion to this five-parter is manifest.	Quia igitur angulus rectus est primi et secundus pars recti evidenter patet utrumque. quare tota liquet haec penthamembris conclusio.
¶ Though from those last two parts also note a proof contrary those two arguments, an example about which we considered in the 15th of this, for it is passing from an angle of the portion less than a semicircle, which is less than right by the last part of this, to an angle of the portion greater than a semicircle, which is greater than right by the penultimate part of this, yet not through one equal [to a right].	¶ Ex istis autem duobus ultimis partibus nota etiam instantiam contra illas duas argumentationes ad quas tulimus instantiam. in .15. huius. transitur enim ab angulo portionis semicirculo minoris qui est minor recto per ultimam partem huius ad angulum portionis semicirculo maioris qui est maior recto per penultimam partem huius. non tamen per aequale.
For that every portion of a circle is a semicircle, or greater than a semicircle, or less, then let the angle of the semicircle by the second part of 15 be as the angle of the smaller portion by the last part of this, less than right, and let that of the greater portion be greater than right, however, it will not be an angle of any portion, nor simply any contained by the circumference and a right line, nor will it be right or equal to right.	Cum enim omnis portio circuli sit semicirculus aut maior semicirculo. aut minor: sit autem tam angulus semicirculi per primam [sic] partem .15. quam angulus portionis minoris per ultimam partem huius minor recto. portionis vero maioris sit maior recto. et tamen non erit alicuius portionis angulus. nec simpliciter aliquis contentus a circumferentia. et linea recta nec rectus nec aequalis recto.
So that it may be more clear, suppose in the circle a.b.c, whose center is d, the line a.b, whose bounds shall have no limit from point b, is dividing from that a portion smaller than the semicircle, and by the final part of this it will be less than right. Let the diameter of this circle be a.d.c, and so the line a.b is imagined to be pivoting upon part a to part c, which so long as it is on that side of c or on c and covering the diameter a.d.c, with the arc it shall make an angle less than right.	Quod ut clarius pateat sit in circulo .a.b.c. cuius centrum .d. linea .a.b. cui non sit determinatus finis ex parte .b. secans ex ipso portionem semicirculo minorem. eritque per ultimam partem huius minor recto. huius circuli sit diameter .a.d.c. et imaginetur linea .a.b. moveri ad partem .c. super punctum .a. quae quamdiu fuerit citra .c. vel in ipso .6. [sic] cooperiens diametrum .a.d.c. faciet cum arcu angulum minorem recto.
But at every point beyond c, such as at e, it will make an angle greater than right by the penultimate of this, and so, it is passing from lesser to greater and not through the equal. And just as with right-lined angles it is possible to find a greater angle of the semicircle and a lesser, yet not one equal, as was demonstrated in the 15th of this and by extension within the angles of a portion, those greater than right and less than may be found, yet not shall be equal, as is clear from this demonstration.	In omni autem puncto ultra .c. velut in .e. faciet per penultimum partem huius angulum maiorem recto. transitur ergo a minori ad maius non per aequale. et sicut in rectilineis angulis est reperire maiorem angulo semicirculi et minorem. non tamen aequalem ut monstratum est. in .15. huius: sic in angulis portionis est reperire maiorem recto et minorem non tamen aequalem: ut patet ex ista demonstratione.
Proposition 31	Propositio .31.
[English] Latin	English [Latin][¶]
If a right line contacts a circle and from the point of contact an other right line is drawn within the circle, dividing the circle off center, whatever two angles it makes at the point of contact are equal to the two angles that are over the arc in alternate portions of the circle.	Si circulum linea recta contingat et a contactu in circulum quaedam circulum secans recta linea praeter centrum ducatur quoscunque duos angulos cum contingente facit. duobus angulis qui in alternatis circuli super arcus consistunt portionibus: aequales sunt.
¶ Suppose the right line a.b is contacting the circle c.d.e.f, whose center is g, at point d, from which the line d.f is drawn off center within the circle, dividing it. And the angle d.c.f is made resting over the arc of the portion d.c.f by drawing lines c.d and c.f, and the angle d.e.f is made resting over the arc of the portion d.e.f by drawing lines e.d and e.f. I say angle c is equal to the angle b.d.f, and angle e to the angle a.d.f. For the line of the diameter d.g.h and the line f.h are drawn, and by the 17th of this a.b is perpendicular to d.h, and by the first part of the preceding the angle d.f.h is right, by which the two angles a.d.h and d.f.h are equal. And given the common angle h.d.f, the angle a.d.f equals the two angles that are d.f.h and h.d.f, but those two with angle h are equal to two rights by the 32nd of the first, therefore, the angle a.d.f and angle h are equal to two rights. But angle a.d.f and the angle b.d.f are equivalent to two rights by the 13th of the first, thus the angle b.d.f is equal to angle h, and so to angle c by the 20th of this, and this is the first. And that the two angles c and e are equal to two rights by the 21st of this, angle e will equal the angle a.d.f, which is the second.	¶ Sit recta linea .a.b. contingens circulum .c.d.e.f. cuius centrum .g. in puncto .d. a quo .d. ducatur in circulum praeter centrum linea .d.f. secans ipsum. fiantque angulus .d.c.f. consistens super arcum portionis .d.e.f.[sic] ductis lineis .c.d. et .c.f. et angulus .d.e.f. consistens super arcum portionis .d.e.f. ductis lineis .e.d. et .e.f. dico angulum .c. esse aequalem angulo .b.d.f. et angulum .e. angulo .a.d.f. ducantur enim diametri .d.g.h. et linea .f.h. eritque per .17. huius .d.h. perpendicularis super .a.b. et per primam partem praemissae angulus .d.f.h. rectus. quare duo anguli .a.d.h. et d.f.h. sunt aequales. posito ergo communi angulo .h.d.f: erit angulus .a.d.f. aequalis duobus angulis qui sunt .d.f.h. et .h.d.f. sed hi duo cum angulo .h. sunt aequales duobus rectis per .32. primi: ergo angulus .a.d.f. cum angulo .h. aequales duobus rectis. sed angulus .a.d.f. cum angulo .b.d.f. aequivalet duobus rectis per .13. primi: ergo angulus .b.d.f. est aequalis angulo .h. ergo et angulo .c. per .20. huius et hoc est primum. et quia duo anguli .c. et .e. sunt aequales duobus rectis per .21. huius erit angulus .e. aequalis angulo .a.d.f. quod est secundum.
Or let this be the second, whereby the angle a.d.f with angle h is equivalent to two rights, as has been demonstrated. But angle e with angle h is equivalent to two rights by the 21st of this, therefore, angle e is equal to angle a.d.f, which is the intent.	Vel istud secundum sit angulus .a.d.f. cum angulo .h. aequivalet duobus rectis. ut praemonstratum est. sed angulus .e. cum angulo .h. aequivalet duobus rectis per .21. huius. ergo angulus .e. est aequalis angulo .a.d.f. quod est propositum.
Proposition 32	Propositio .32.
[English] Latin	English [Latin][¶]
Over a given line, to describe a portion of a circle taking an angle equal to a given angle, be it either right or greater or less than right.	Super datam lineam. circuli portionem describere capientem angulum. dato angulum aequalem. seu rectum. seu maiorem seu minorem recto.
¶ Let a.b be the given line and c the given angle. Over the line a.b I wish to draw one portion of a circle, admitting at the circumference a right-lined angle equal to angle c.	¶ Sit .a.b. linea data et .c. datus angulus super lineam .a.b. volo describere unam circuli portionem recipientem in circumferentia rectilineum angulum aequalem angulo .c.
And if the angle is right, a.b is divided through the middle, over which I shall draw a semicircle and the deed will then be done by the first part of the 30th of this.	Si igitur fuerit angulus .c. rectus divisa .a.b. per medium describam super eam semicirculum. factumque erit propositum. per primam partem .30. huius.
¶ But if it were obtuse, I might draw the line d.a with the line b.a containing an angle equal to the angle c. So from point a I will draw the line a.e perpendicular to the line a.d, and through the 23rd of the first make an angle upon point b equal to the angle e.a.b, by which the obtuse will exceed right by drawing the line b.f up until the perpendicular a.e, and by the 6th of the first the lines f.a and f.b will be equal. Then making point f the center of a circle, I will draw the circle a.h.b following the extent of the line f.a, and by a corollary of the 15th of this, the line a.d is touching the circle, whereby through the preceding the angle that is on the portion a.h.b is equal to the angle d.a.b and so to angle c, which is the intent.	¶ Si autem sit obtusus ducam lineam .d.a. cum linea .b.a. continentem aequalem angulum angulo .c. et a puncto .a. ducam lineam .a.e. perpendicularem super lineam .a.d. et super punctum .b. faciam angulum per .23. primi aequalem angulo .e.a.b. in quo obtusus excedet rectum. ducta linea .b.f. usque ad perpendicularem .a.e. eruntque per .6. primi: lineae .f.a. et .f.b. aequales: facto itaque puncto .f. centro circuli describam secundum quantitatem lineae .f.a: circulum .a.h.b. eritque per correlarium .15. huius linea .a.d. contingens circulum. quare per praemissam angulus qui sit in portione .a.h.b. est aequlis angulo .d.a.b. quare et angulo .c. quod est propositum.
¶ But if angle c were acute, I might draw the line a.g with the line a.b containing an angle equal to angle c. So from point a I will draw a.e perpendicular to the line a.g, and upon point b make an angle equal to the angle e.a.b by which right shall exceed the acute by drawing the line b.f up until the perpendicular line a.e, and by the 6th of the first the lines f.a and f.b are equal. Then making point f the center of a circle, I will draw the circle a.k.b following the extent of the line f.a and by a corollary of the 15th of this, the line a.g is contacting the circle, whereby through the preceding the angle that is on the portion a.k.b is equal to the angle g.a.b and so to angle c, which is the intent.	¶ Si autem angulus .c. sit acutus. producam lineam .a.g: continentem cum linea .a.b. angulum aequalem angulo .c. et a puncto .a. ducam .a.e. perpendicularem ad lineam .a.g. et super punctum .b. faciam angulum aequalem angulo .e.a.b. in quo rectus excedit acutum. ducta linea .b.f. usque ad perpendicularem .a.e. eruntque per .6. primi lineae .f.a. et .f.b. aequales: facto itaque puncto .f. centro circuli. describam secundum quantitatem lineae .f.a. circulum .a.k.b. eritque per correlarium .15. huius linea .a.g. contingens circulum. quare per praemissam angulus qui sit in portione .a.k.b. est aequalis angulo .g.a.b. quare et angulo .c. quod est propositum.
Proposition 33	Propositio .33.
[English] Latin	English [Latin][¶]
From a given circle, to abscind a portion taking an angle equal to a given angle.	A dato circulo: dato angulo. aequum angulum capientem portionem abscindere.
¶ Let a.b be the given circle and c the given angle, and accordingly I wish to abscind one portion from the circle a.b taking an angle equal to the angle c. I draw out the line d.a.e contacting the given circle at point a, from which I draw the line a.b within the circle that, with the line a.e, shall contain an angle equal to the angle c, and by the 31st of this the portion a.b is on the side of the line a.d, which is admitting an angle equal to angle c, which is the intent.	¶ Sit .a.b. datus circulus. et .c. datus angulus. volo ergo a circulo .a.b. abscindere portionem unam capientem aequalem angulum angulo .c. produco lineam .d.a.e. contingentem datum circulum in puncto .a. a quo ducto in circulum lineam .a.b. contintem cum linea .a.e. angulum aequalem angulo .c. eritque per .31 huius portio .a.b. existens a parte lineae .a.d. recipiens angulum aequalem angulo .c. quod est propositum.
Proposition 34	Propositio .34.
[English] Latin	English [Latin][¶]
If in a circle two right lines cut one another, that which proceeds within the two parts of one of them is equal to the rectangle that is contained within the two parts of the other line.	Si intra circulum duae rectae lineae sese invicem secant. quod sub duabus partibus unius earum procedit. aequum est ei rectangulo quod sub duabus alterius lineae partibus continetur.
¶ Suppose the two lines, a.c and b.d, cut one another over point e in the circle a.b.c.d. I say that the rectangle that is produced from a.e according to e.c is equal to that which is produced from b.e according to e.d. For either both lines, a.c and b.d, will pass through the center of the circle, or only one, or neither, and if both were to pass through the center, e will be the center of the circle and all 4 lines will be equal and so the proposition is clear.	¶ Sint duae lineae .a.c. et .b.d. secantes se in circulo .a.b.c.d. super punctum .e. dico quod illud rectangulum quod fit ex .a.e. in .e.c. aequum est ei quod fit ex .b.e. in .e.d. aut enim ambae lineae .a.c. et .b.d. transibunt per centrum circuli aut altera tantum aut neutra. quod si ambae transeant per centrum. erit .e. centrum circuli. omnesque .4. lineae aequales: quare liquet propositum.
¶ But if only one of them passes through the center, suppose that is b.d and the center of the circle is f, and either b.d will divide a.c by equals or by unequals, then let it divide first by equals, and it will, by the first part of the 3rd of this, be dividing it orthogonally. And the line f.c is drawn and by the 5th of the second, that produced from b.e according to e.d with the quadrate e.f is equal to the quadrate of the line f.d, and so to the quadrate of the line f.c, and so by the penultimate of the first to the quadrates of the two lines f.e and e.c. Then by removing the quadrate e.f from both, that produced from b.e according to e.d will equal the quadrate of the line e.c, and because e.c is equal to a.e the proposition is clear.	¶ Quod si altera earum tantum transit per centrum sit illa .b.d. centrumque circuli sit .f. aut ergo .b.d. secabit .a.c. per aequalia aut per inaequalia. secet ergo primo per aequalia: eritque per primam partem .3. huius secans eam orthogonaliter. ducatur itaque linea .f.c. eritque per .5. secundi quod fit ex .b.e. in .e.d. cum quadrato .e.f. aequale quadrato lineae .f.d. quare et quadrato lineae .f.c. ergo per penultimam primi et quadratis duarum linearum .f.e. et .e.c. dempto ergo utrique quadrato .e.f. erit quod fit ex .b.e. in .e.d. aequale quadrato lineae .e.c. et quia .e.c. est aequalis .a.e. patet propositum.
¶ But if b.d passes through the center dividing a.c by unequals, from the center f, f.g is drawn perpendicular to a.c and by the second part of the third of this a.g will be equal to g.c. And the line f.c is drawn and by the 5th of the second, that produced from b.e according to e.d with the quadrate e.f (and by the penultimate of the first with the quadrates of the two lines f.g and g.e, for that the angle f.g.e is equal to right) is equal to the quadrate of the line d.f, and so of the line f.c, and for that by the penultimate of the first, to the quadrates of the two lines f.g and g.c. Then by removing the quadrate of the line f.g from both, that produced from b.e according to e.d with the quadrate of the line g.e will equal the quadrate of the line g.c, but by the 5th of the second, that produced from a.e according to e.c with the quadrate of the line g.e is equal to that produced from the same quadrate g.c. Therefore, by removing the quadrate of the line g.e from both, that produced from b.e according to e.d will equal that produced from a.e according to e.c, which is the intent.	¶ Quod si .b.d. transiens per centrum secat .a.c. per inaequalia a centro .f. ducatur .f.g. perpendicularis ad .a.c. eritque per secundam partem tertii huius .a.g. aequalis .g.c. et ducatur linea .f.c. eritque per .5. secundi quod fit ex .b.e. in .e.d. cum quadrato .e.f.c. [sic] et ideo per penultimam primi cum quadratis duarum linearum .f.g. et .g.e. propter id quod angulus .f.g.e. est recto et est aequalis quadrato lineae .d.f. et ideo lineae .f.c. propter quod per penultimam primi et quadratis duarum linearum .f.g. et .g.c. dempto ergo utrique quadrato lineae .f.g. erit quod fit ex .b.e. in e.d. cum quadrato lineae .g.e. aequale quadrato lineae .g.c. sed per .5. secundi quod fit ex .a.e. in .e.c. cum quadrato lineae .g.e. est aequum ei quod fit ex .a.e. in .e.c. cum quadrato eiusdem .g.e. [sic]. dempto igitur utrique quadrato lineae .g.e: erit quod fit ex .b.e. in .e.d. aequale ei quod fit ex .a.e. in .e.c. quod est propositum.
¶ But if neither of them pass through the center, one shall divide the other either by equals or by unequals, and I will draw the line g.f.e.h through the diameter of the circle, passing through the point of their division. And if one divides the other bye equals, as b.d and a.c, then g.h shall also divide a.c by equals, and so it is orthogonal by the third of this, therefore, by the second measure of this conclusion, that produced from g.e according to e.h is equal to that produced from a.e according to e.c, and by the third measure of this, that produced from g.e according to e.h is equal to that produced from b.e according to e.d, and so that produced from a.e according to e.c is equal to that produced from b.e according to e.d, which is the intent.	¶ Quod si neutra earum transit per centrum sive altera dividat alteram per aequalia sive per inaequalia. producam lineam .g.f.e.h. diametrum circuli transeuntem per punctum sectionis earum. Et si altera dividat alteram per aequalia. ut .b.d.a.c. tunc .g.h. dividit etiam .a.c. per aequalia. ergo orthogonaliter per tertiam huius. ergo per secundum modum huius conclusionis quod fit ex .g.e. in .e.h. aequum est ei quod fit ex .a.e. in .e.c. et per tertium modium huius quod fit ex .g.e. in .e.h. aequum est ei quod fit ex: b.e. in e.d. ergo quod fit ex .a.e. in .e.c. aequum est ei quod fit ex .b.e. in .e.d. quod est propositum.
¶ And moreover, if neither divides the other by equals, by the third measure of this conclusion that produced from g.e according to e.h will equal both of those that are produced from a.e according to e.c and b.e according to e.d, whereby one of them will equal the other, which is the intent.	¶ At si neutra dividit alteram per aequalia erit per tertium modum huius conclusionis quod fit ex .g.e. in .e.h. aequale utrique eorum quae fiunt ex .a.e. in .e.c. et .b.e. in .e.d. quare unum eorum erit aequale alteri: quod est propositum.
Proposition 35	Propositio .35.
[English] Latin	English [Latin][¶]
If a point is marked without a circle whence two right lines are drawn to the circle, with one line cutting and the other contacting, then that contained within the whole cutting line as well the extrinsic part of it, is equal to the quadrate that is drawn from the contacting line.	Si extra circulum punctus signetur. ab eo autem ad circulum alia linea secans. alia contingens duae rectae lineae ducantur quod sub tota secante. atque parte sui extrinseca continetur aequum est ei quadrato quod ex contingente linea describitur.
¶ Let a be the point marked without the circle b.c.d, whose center is e, wherefrom two lines are drawn to the circle, a.b contacting it and a.c.d cutting. I say that that which is produced from a.c according to d.a is equal to the quadrate of the line a.b, for either a.d.c will pass through the center or it will not, therefore, first let it pass through the center, which is e, and the line e.b shall be drawn, which by the 17th of this will be perpendicular to the line a.b. And that the line d.c is divided by equals at point e and the line d.a is added to it, by the sixth of the second that produced from c.a according to a.d with the quadrate of the line e.d, and thereby the quadrate of the line e.b, will equal the quadrate of the line e.a, and so by the penultimate of the first it is equal the quadrates of the two lines e.b and b.a, for that angle b is right. Then by removing the quadrate e.b from each, that produced from c.a according to a.d will equal the quadrate of the line a.b, which is the intent.	¶ Sit .a. punctus signatus extra circulum .b.c.d. cuius centrum .e. a quo ducantur ad circulum duae lineae .a.b. contingens. et .a.c.d. secans. dico quod illud quod fit ex .a.c. in .d.a. aequum est quadrato lineae .a.b. aut enim .a.d.c. transit per centrum aut non transeat: ergo primo per centrum quod est .e. et ducatur linea .e.b. quae per .17. huius perpendicularis erit super lineam .a.b. et quia linea .d.c. divisa est per aequalia in puncto .e. et est ei addita linea .d.a. erit per sextam secundi quod fit ex .c.a. et .a.d. cum quadrato lineae .e.d. et ideo cum quadrato lineae .e.b. aequale quadrato lineae .e.a. et ideo per penultimam primi aequale quadratis duarum linearum: e.b. et .b.a. propter id quod angulus .b. rectus. dempto ergo utrimque quadrato .e.b. erit quod fit ex .c.a. in .a.d. aequale quadrato lineae .a.b. quod est propositum.
¶ But if the line a.d.c does not pass through the center, then a.f.e.g is presupposed to pass through the center and the lines e.d and e.h are drawn. And say e.h is perpendicular to a.d.c, and by the 3rd of this d.h equals h.c for that the line d.c is divided by equals at point h. And the line a.d is added to it, so by the 6th of the second that produced from c.a according to a.d with the quadrate d.h equals the quadrate of the line a.h, and so when you add the quadrate h.e to both, that produced from c.a according to a.d with the quadrates of the two lines d.h and h.e, and thereby the penultimate of the first with the quadrate d.e for that angle h is right, and so with the quadrate e.f, for that e.d and e.f are equal, will equal the quadrates of the two lines a.h and h.e, and thereby the penultimate of the first the quadrate of the line a.e. But by the sixth of the second, that produced from g.a according to a.f with the quadrate f.e is equal to the quadrate of the line a.e, and so because each of those, which are produced from c.a according to a.d and from g.a according to a.f, with the quadrate of the line f.e, are equal to the quadrate of the line a.e, they themselves will be equal to one another.	¶ Quod si linea .a.d.c. non transit per centrum sumatur .a.f.e.g. transiens per centrum et ducantur lineae .e.d. et .e.h. et sit .e.h. perpendicularis ad .a.d.c. eritque per .3. huius d.h. aequalis .h.c. quia ergo linea .d.c. divisa est per aequalia in puncto .h. et addita sibi linea .a.d. erit per .6. secundi quod fit ex .c.a. in .a.d. cum quadrato .d.h. aequale quadrato lineae .a.h. ergo addito utrique quadrato .h.e. erit quod fit ex .c.a. in .a.d. cum quadratis duarum linearum .d.h. et .h.e. et ideo per penultimam primi cum quadrato .d.e. propter id quod angulus .h. est rectus. et ideo cum quadrato .e.f. propter id quod .e.d. et .e.f. sunt aequales. aequale quadratis duarum linearum .a.h. et .h.e. et ideo per penultimam primi quadrato lineae .a.e. sed quia per sextam secundi quod fit ex .g.a. in .a.f. cum quadrato .f.e. aequale est quadrato lineae .a:e. quia ergo utrimque eorum quae fiunt ex .c.a. in .a.d. et ex .g.a. in .a.f. cum quadrato lineae .f.e. est aequale quadrato lineae .a.e. ipsa erunt inter se aequalia.
Therefore, by removing both from the quadrate of the line e.f, that produced from c.a according to a.d will equal that produced from g.a according to a.f. But that that produced from g.a according to a.f is equal to the quadrate of the line a.b by the preceding method of this, then that produced from c.a according to a.d. is equal to the quadrate a.b, which is the intent.	Dempto ergo utrinque quadrato lineae .e.f. erit quod fit ex .c.a. in .a.d. aequale ei quod fit ex .g.a. in .a.f. sed id quod fit ex .g.a. in .a.f. est aequale quadrato lineae .a.b. per praemissum modum huius: ergo quod fit ex .c.a. in .a.d. est aequale quadrato lineae .a.b. quod est propositum.
¶ And from this note, that from a point marked without a circle, if any cutting lines are drawn from it to the circle then the rectangles that are contained within the wholes of them and their extrinsic parts are mutually equal, for that all are equal to the quadrate of the contacting line.	¶ Et ex haec nota quod puncto extra circulum signato si ab ipso ad circulum quolibet secantes lineae ducantur rectangula quae continentur sub totis. et earum portionibus extrinsecis adinvicem sunt aequalia. quoniam omnia sunt aequalia quadrato lineae contingentis.
¶ Also note that if from a point marked without the circle any two contacting lines are drawn to the circle, they are mutually equal, for the quadrate of each of them will equal that which is produced by the cutting line from the point itself to that drawn into the circle and according to its extrinsic part, and this is moreover evidently manifest by the penultimate of the first.	¶ Nota etiam quod si a quolibet punto extra circulum signato duae lineae contingentes ad circulum ipsum ducantur. ipse erunt adinvicem aequales. erit enim quadratum utriusque earum aequale ei quod fit ex linea secante ab ipso puncto ducta in circulum. in partem eius extrinsecam. hoc autem evidentius patet per penultimam primi.
¶ Suppose a is the point marked without the circle b.c.d, whose center is e, and from it the two lines a.b and a.d are drawn, contacting the circle at points b and d. I say that these are to be equal. For I will produce the lines e.a, e.b, and e.d, and by the 17th of this both of the angles b and d are right, wherefore by the penultimate of the first, the quadrate for a.c is equal to the two quadrates of the two lines a.b and b.e.	¶ Sit .a. punctus signaus extra circulum .b.c.d. cuius centrum .e. et ab ipso ducantur duae lineae .a.b. et .a.d. contingentes circulum in punctis .b.d. dico ipsas esse aequales. producam enim lineas .e.a. e.b. et .e.d. eritque per .17. huius uterque angulorum .b. et .d. rectus. quare per penultimam primi quadratum .a.c. aequale duobus quadratis duarum linearum .a.b. et .b.e.
And similarly as well from the two of the two a.d and d.e, wherefore the quadrate of the two lines, a.b and b.e are equal to the quadrate of the two lines a.d and d.e, and because the quadrates of the two that are b.e and e.d are equal, the quadrates of the two that are a.b and a.d will be equal, thus a.b is equal to a.d, which is the intent.	Similiter quoque et duobus: duarum .a.d. et .d.e. quare quadrata duarum linearum a.b. et .b.e. sunt aequalia quadratis duarum .a.d. et .d.e. et quia quadrata duarum quae sunt .b.e. et .e.d. sunt aequalia: erunt quadrata duarum quae sunt .a.b. et .a.d. aequalia: ergo .a.b. est aequalis .a.d. quod est propositum.
¶ And another thing. Say the line b.d is drawn so by the fifth of the first, the angle e.b.d equals the angle e.d.b, for that the line e.b is equal to the line e.d, and because each of the two angles b and d are right, through common knowledge the remaining angle a.b.d equals the remaining angle a.d.b, and so by the sixth of the first, the line ab is equal to the line a.d.	¶ Aliter etiam ducatur linea .b.d. eritque per quintam primi angulus .e.b.d. aequalis angulo .e.d.b. propter id quod linea .e.b. est aequalis lineae .e.d. et quia uterque duorum angulorum .b. et .d. est rectus. erit per communem scientiam angulus .a.b.d. residuus aequalis angulo .a.d.b. residuo. per sextam. ergo primi est linea .a.b. aequalis lineae .a.d.
Proposition 36	Propositio .36.
[English] Latin	English [Latin][¶]
If a point is marked without a circle whence two lines are drawn to the circumference, one cutting and the other applied to the circumference, and that produced from the whole of the cutting line drawn according to the extrinsic part of it is equal to that produced from the applied line drawn according to itself, from necessity the applied line will be touching the circle.	Si fuerit punctus extra circulum signatus a quo duae lineae ad circumferentam ducantur. altera secans. altera circumferintiae applicata fueritque quod ex ductu totius secantis in partem sui extrinsecam aequum ei quod ex ductu applicatae in seipsam fit: erit linea applicata ex necessitate circulum contingens.
¶ Let a be the point marked without the circle b.c.d, whose center is e, wherefrom the line a.b.d is drawn to the circle, cutting it, and the line a.c is drawn, applied to the circumference, and that that produced from d.a according to a.b equals the quadrate a.c. I say the line a.c is contacting, and this is a converse of the preceding.	¶ Sit a punctus signatus extra circulum .b.c.d. cuius centrum .e. a quo ducantur ad circulum linea .a.b.d. secans ipsum et linea .a.c. applicata circumferentiae. et esto ut quod fit ex .d.a. in a.b. sit aequale quadrato .a.c. dico lineam .a.c. esse contingentem et est haec conversa prioris.
For if it is not contacting, then suppose the line a.f is contacting, and by the preceding that produced from d.a according to a.b will equal the quadrate of the line a.f, hence the quadrate of the line a.f is equal to the quadrate of the line a.c, and so a.c is equal to a.f, which is impossible by the 8th of this. And so a.c is contacting, which is the intent.	Si enim non est contingens. sit ergo contingens linea a.f: eritque per praemissam quod fit ex .d.a. in .a.b. aequale quadrato lineae .a.f. quare quadratum lineae. a.f. est aequale quadrato lineae .a.c. ergo .a.c est aequalis .a.f. quod est impossibile. per .8. huius. erit ergo .a.c. contingens quod est propositum.
¶ The same of the ostensive will be demonstrated, for the prior could apply to the disposed as well the hypothesis. And if the line a.b.d passes through the center the line c.e is drawn, which by the 6th of the second, that produced from d.a according to a.b with the quadrate e.b, and so with the quadrate e.c, is equal to the quadrate a.e. But that produced from d.a according to a.b is posited equal to the quadrate a.c, therefore, the quadrate a.c with the quadrate c.e is equal to the quadrate a.e, and by the last of the first, angle c is right. Thus, by a corollary of the 15th of this, the line a.c is contacting the circle, which is the intent.	¶ Idem ostensivae probabitur maneat prior disposito et ypothesis. et si linea .a.b.d. transit per centrum ducatur linea .c.e. quae erit per .6. secundi quod fit ex .d.a. in .a.b. cum quadrato .e.b. et ideo cum quadrato .e.c. aequale quadrato .a.e. sed quod fit ex .d.a. in .a.b. positum est aequale quadrato .a.c. ergo quadratum .a.c. cum quadrato .c.e. est aequale quadrato .a.e. ergo per ultimam primi angulus .c. est rectus. ergo per correlarium 15. huius linea .a.c. est contingens circulum quod est propositum.
¶ And if a.b.d does not pass through the center, then from point a, a line is drawn passing through the center, and because that made thus from the whole according to its extrinsic part is equal to that produced from d.a according to a.b, by the preceding that will equal the quadrate of the line a.c, and just as before, a.c will be contacting the circle.	¶ Si autem .a.b.d. non transit per centrum ducatur a puncto .d. [sic] linea transiens per centrum. et quia quod fit ex hac tota in eius partem extrinsecam est aequale ei quod fit ex .d.a. in .a.b. per praemissam ipsum erit aequale quadrato lineae .a.c. quare ut prius .a.c. erit contingens ciruculum.
¶ The third book ends.	¶ Explicit liber tertius.
¶ The third book begins.	¶ Incipit liber tertius.

[English] Latin	English [Latin][¶]
¶ A figure within a figure is said to be inscribed when that which is inscribed is of that in which it is inscribed, and the sides contain every one of the inscribed angles within the interior part.	¶ Figura intra figuram dicitur inscribi quando ea quae inscribitur eius in qua inscribitur. latera uno quoque suorum angulorum ab interiore parte contingit.
¶ But the figure of a figure is said to be circumscribed whenever the very figure which is circumscribed contains every angle of that within all its sides.	¶ Circumscribi vero figura figurae perhibetur quotiens ea quidem figura eius cui circumscribitur suis lateribus omnibus omnes angulos contingit.
Proposition 1	Propositio .1.
[English] Latin	English [Latin][¶]
Within a given circle, to fit a right line equal to a given right line that is no greater than the diameter.	Intra datum circulum datae lineae rectae quae diametro minime maior existat aequam rectam lineam coaptare.
¶ Suppose the given line is a.b and the given circle c.d.e, whose diameter is c.d, which is not greater than the line a.b. I want to fit a line equal to a.b within the given circle which, if it were equal to the diameter, then the intention is certain. But if d.f is assumed less than the diameter, then over point d the circle f.e.g is drawn following a length equal to it, cutting the given circle at points g and e, and a line is drawn to one of them from point d, as d.e or d.g, and either the two of them will equal the line a.b and so both of them are equal to the line d.f by the definition of the circle, whereby we have the intention.	¶ Sit linea data .a.b. circulusque datus .c.d.e. cuius diameter .c.d. qua non est maior linea .a.b. volo intra datum circulum coaptare lineam aequalem .a.b. quae si fuerit aequalis diametro constat propositum. si autem minor ex diametro sumatur .d.f. sibi aequalis et super punctum .d. secundum quantitatem lineae .d.f. describatur circulus .f.e.g. secans datum circulum in punctis .g. et .e. ad alterum quorum ducatur linea a puncto .d. ut .d.e. vel .d.g. eritque utralibet earum aequalis lineae .a.b. eo quod utraque earum est aequalis lineae .d.f. per diffinitionem circuli: quare habemus propositum.
Proposition 2	Propositio .2.
[English] Latin	English [Latin][¶]
Within an assigned circle, to assemble a triangle equiangular to an assigned triangle.	Intra assignatum circulum triangulum triangulo assignato aequiangulum collocare.
¶ Suppose the assigned triangle is a.b.c and the assigned circle d.e.f. Within this circle I want to assemble one triangle equiangular to the triangle a.b.c, and it is not necessary for it to be equilateral, though it is possible. I draw g.d.h contacting the circle at point d, upon which I produce angle h.d.f with the drawn line d.f, equal to angle c, and angle g.d.e with the drawn line d.e, equal to angle b, and I extend the line e.f. And by the 33rd of the third angle e will equal angle c since each is equal to the angle h.d.f, thus c, by position, shall indeed equal e by the 31st of the third. By the same rationale, angle f will equal angle b and so by the 32nd of the first, the third d will equal the third a, whereby we have the intention.	¶ Sit assignatus triangulus .a.b.c. assignatusque circulus .d.e.f. volo intra hunc circulum collocare unum triangulum aequiangulum triangulo .a.b.c. aequilaterum enim non est necessarium esse sed est possibile. produco .g.d.h. contingentem circulum in puncto .d. super quem facio angulum .h.d.f. ducta linea .d.f. aequalem angulo .c. et angulum .g.d.e. ducta linea .d.e. aequalem angulo .b. et protraho lineam .e.f. eritque per .31. [sic] tertii angulus .e. aequalis angulo .c. quia uterque est aequalis angulo .h.d.f.c. quidem per positionem .e. vero per .31. tertii eadem ratione erit angulus .f. aequalis angulo .b. quare per .32. primi .d. tertius erit aequalis .a. tertio. quare habemus propositum.
Proposition 3	Propositio .3.
[English] Latin	English [Latin][¶]
Around an assigned circle, to describe a triangle equiangular to an assigned triangle.	Circa assignatum circulum assignato triangulo triangulum aequiangulum describere.
¶ Suppose as before the assigned triangle is a.b.c and the assigned circle d.e.f, whose center is g. Around this circle I want to describe one triangle equiangular to the triangle a.b.c, and it is not necessary that it is equilateral, though it is possible. I will lengthen the base b.c at both ends so that two extrinsic angles are produced, and from the center g I will draw the line g.d to the circumference, and construct angle d.g.e with the drawn line g.e, equal to angle b from without, and d.g.f with the drawn line g.f, equal to c from without, and from points d, e, and f, I will draw lines orthogonally in each direction that by a corollary of the 15th of the third shall contact the circle, and those lines I will extend until they meet at points h, k, and l. And it is necessary they are to meet, for since each of the angles that are at d and each of those that are at e shall be right, if it is intended for the line d.e to be lengthened, then the two angles that are at part h will be less than two rights, and so by the penultimate petition lines l.d.h and k.e.h will meet at that side of the protraction. By the same rationale, the two lines h.d.l and k.f.l will meet when each of the angles that are at f are also right.	¶ Sint ut prius assignatus triangulus .a.b.c. assignatusque circulus d.e.f. cuius centrum .g. circa hunc circulum volo describere unum triangulum aequiangulum triangulo .a.b.c. aequilaterum enim non est necessarium sed est possibile. producam basim .b.c. in utramque partem. ut fiant duo anguli extrinseci. et a centro .g. producam lineam .g.d. ad circumferentiam. et constituam angulum .d.g.e. ducta linea .g.e. aequalem angulo .b. extrinseco. et .d.g.f. ducta linea .g.f. aequalem .c. extrinseco. et a punctis .d.e.f producam in utranque partem lineas orthogonaliter quae per correlarium .15. tertii erunt contingentes circulum quas contingentes protraham quousque concurrant in punctis .h.k.l. neccesse est enim ipsas concurrere. cum enim uterque angulorum qui sunt ad .d. et uterque eorum qui sunt ad .e. sit rectus si intelligatur protrahi linea .d.e. erunt duo anguli qui sunt ad partem .h. minores duobus rectis. quare per penultimam petitionem in partem illam protractae concurrent lineae .l.d.h.k.e.h. eadem ratione concurrent duae lineae .h.d.l.k.f.l. cum uterque angulorum qui sunt. ad .f. sit etiam rectus.
Since, then, the two angles d and e in the quadrilateral h.d.e.g are right, the two angles g and h will equal two right angles because the four angles of any quadrilateral are equal to four rights, as is previously demonstrated by the 32nd of the first. And because the two angles at b, within and without, are likewise equal to two rights by the 13th of the first, and moreover, the extrinsically placed is equal to d.g.e, then b from without will equal h. And by the same rationale c from within will equal l, and because the two angles b and c from within are less than two rights by the 17th of the first, the two angles h and l will also be less than two rights, whereby through the penultimate petition the two extended lines, h.e and l.f, meet at point k, and so the triangle h.k.l is made. And since angle h is equal to angle b from within and angle l to angle c from within, by the 32nd of the first, angle k will equal angle a, thus we have the intention.	Quia ergo in quadrilatero .h.d.e.g. duo anguli .d. et .e. sunt recti. erunt duo anguli .g. et .h. aequales duobus rectis. cuiuslibet enim quadrilateri quatuor anguli sunt aequales quatuor rectis. ut monstratum est supra .32. primi: et quia duo anguli .b. intrinsecus et extrinsecus sunt similiter aequales duobus rectis per .13. primi: at vero .b. extrinsecus positus est aequalis .d.g.e. erit intrinsecus .b. aequalis .h. simili quoque ratione erit .c. intrinsecus aequalis .l. et quia duo anguli .b. et c. intrinsecui sunt minores duobus rectis. per .32. [sic] primi: erunt similiter duo anguli .h. et .l. minores duobus rectis. quare per penultimam petitionem duae lineae .h.e. et .l.f. protractae concurrent in puncto .k. fietque triangulus .h.k.l. et quia angulus .h. est aequalis angulo .b. intrinseco. et angulus .l. angulo .c. intrinseco. erit per .32. primi: angulus .k. aequalis angulo .a. quare habemus propositum.
Proposition 4	Propositio .4.
[English] Latin	English [Latin][¶]
To describe a circle within a given triangle.	Intra datum triangulum circulum describere.
¶ Suppose the assigned triangle is a.b.c, wherewithin I want to describe a circle. This is as though a conversion of the second, for I divide two of its angles, a and b, by equals, that is, a by the drawn line a.d and b with the drawn line d.b, which shall meet at point d, from where I shall draw perpendiculars to the three sides of the triangle, indeed, d.e to a.b, d.f to b.c, and d.g to a.c. And for that of the two triangles, e.a.d and g.a.d, the angle a of one is equal to angle a of the other, and since both of the angles e and g are right, and the side a.d is common, then by the 26th of the first the line d.e will equal the line d.g. By the same rationale, when of the two triangles, e.b.d and f.b.d, the angle b of one shall equal angle b of the other and both of the angles e and f are right, and the side d.b is common, then by the same, the line e.d will equal the line d.f, thus the three lines, d.e, d.f, and d.g, are equal. And so when placing the center at d with the circle described according to the extent of one of them, then by the 9th of the third it will traverse the bounds of the remaining two, and for that by a corollary of the 15th of the third each of the lines a.b, b.c, and c.a, will contact the circle, it is evident the intention has been achieved.	¶ Sit assignatus triangulus .a.b.c. volo intra ipsum circulum describere haec est quasi conversa secundae. divido enim duos eius angulos .a. et .b. per aequalia .a. quidam ducta linea .a.d.b. vero ducta linea .b.d. quae concurrant in puncto .d. a quo ducam perpendiculares ad tria latera ipsius .d.e. quidem: ad .a.b.d.f. ad .b.c. et .d.g. ad .a.c. et quia duorum triangulorum .e.a.d. et .g.a.d. angulus .a. unius est aequalis angulo .a. alterius. et uterque angulorum .e. et .g. rectus et latus .a.d. commune. erit per .26. primi: linea .d.e. aequalis lineae .d.g. eadem ratione cum duorum triangulorum .e.b.d. et .f.b.d. angulus .b. unius sit aequalis angulo .b. alterius et uterque angulorum .e. et .f. rectus: latus quoque .d.b. commune: erit per eandem. linea .e.d. aequalis lineae .d.f. quare tres lineae .d.e.d.f.d.g. sunt aequalis. posito ergo centro in .d. et descripto circulo secundum quantitatem unius earum transibit per .9. tertii per reliquarum duarum extremitates: et quia per correlarium .15. tertii unaquaeque linearum .a.b.b.c. et .c.a. erit contingens circulum. patet perfectum esse propositum.
Proposition 5	Propositio .5.
[English] Latin	English [Latin][¶]
To describe a circle around an assigned triangle, whether it may be orthogonal, amblygonal, or oxygonal.	Circa trigonum assignatum sive illud sit orthogonium sive ambligonium. sive oxigonium circulum describere.
¶ Suppose the assigned triangle is a.b.c, around which I want to describe a circle. This is as though a conversion of the third. I divide two of its sides, a.b and a.c, by equals, that is, a.b at point d and a.c at point e, from where I draw out perpendiculars to the lines a.b and a.c, which I extend until they meet at point f. And d.f and e.f shall meet for that when each of the two angles d and e are right, if it were intended for line d.e to be extended then the two angles at either end to which they are extended would be less than two rights, whereby they will meet by the penultimate petition. Thus, from point f, which is the convergence point that I say is to be the center of the desired circle, I extend lines to the separate angles, which are f.a, f.b, and f.c. And since within the triangle a.d.f the two sides a.d and d.f are equal to the two sides b.d and d.f of the triangle b.d.f, and the angle d of one equals angle d of the other for that they are both right, then by the fourth of the first f.a will equal f.b. By the same rationale f.a will equal f.c through the shared sides and angles of the two triangles a.e.f and c.e.f, therefore, by the 9th of the third point f will be the center of the desired circle. This is a universal demonstration for all types of triangles.	¶ Sit trigonus assignatus .a.b.c. volo circa ipsum describere circulum haec est quasi conversa tertiae. divido duo eius latera .a.b. et .a.c. per aequalia .a.b. quidem in puncto .d. et .a.c. in puncto .e. a quibus punctis produco perpendiculares ad lineas .a.b. et .a.c. quas protraho quousque concurrant in puncto .f. sintque .d.f. et .e.f. concurrent enim quoniam cum uterque angulorum .d. et .e. sit rectus si intelligatur protrahi linea .d.e. fient duo anguli ad partem in quam protrahuntur minores duobus rectis: quare concurrent per penultimam petitionem igitur a puncto .f. qui est punctus concursus quem dico esse centrum circuli quaesiti. protraho lineas ad singulas angulos quae sunt .f.a.f.b.f.c. et quia in triangulo .a.d.f. duo latera .a.d. et .d.f. sunt aequalia duobus lateribus .b.d. et .d.f. trianguli .b.d.f. et angulus .d. unius angulo .d. alterius: quia uterque rectus: erit per quartam primi .f.a. aequalis .f.b. eadem ratione erit .f.a. aequalis .f.c. conpartis lateribus et angulis duorum triangulorum .a.e.f. et .c.e.f. ergo per .9. tertii punctum .f. erit centrum circuli quaesiti. haec is universalis demonstratio ad omnes species trigoni.
¶ However, because the author seems to want to vary the means through disjunction between orthogonal, amblygonal, and oxygonal, each of them is singly demonstrated.	¶ Quia tamen auctor videtur velle medium variare disiungendo inter orthogonium ambligonium et oxigonium. de quolibet eorum singillatim est demonstrandum.
¶ Suppose then the proposed triangle is orthogonal and angle a is right. I divide the side b.c respecting this right angle by equals at f, and from this point that I say is to be the center of the circle to the midpoint of either the remaining sides, which shall be d, I draw the line f.d. And because the line f.d divides the two sides a.b and b.c of the triangle a.b.c by equals, it will be equidistant to the third, namely, the line a.c, for this is previously demonstrated in the 39th of the first. And that angle a is presupposed right, by the second part and by the third of the 29th of the first, both of the angles that are at d are right. The line f.a is thus drawn and by the fourth of the first the line a.f equals the line b.f when the sides and angles of the triangles a.d.f and b.d.f are mutually compared, and because the line b.f is equal to the line c.f, the 3 lines, b.f, a.f, and c.f, are equal to one another, wherefore by the 9th of the third f will be the center of the desired circle.	¶ Sit ergo trigonus propositus orthogonius sitque angulus .a. rectus: latus .b.c. respiciens hunc angulum rectum divido per aequalia in .f. a quo puncto quae dico esse centrum circuli ad medium punctum utriusque duorum reliquorum laterum qui sit .d. duco lineam .f.d. et quia linea .f.d. dividit duo latera .a.b. et .b.c. trianguli .a.b.c. per aequalia: ipsa erit aequidistans tertio. videlicet lineae .a.c. hoc enim demonstratum est supra .39. primi: et quia angulus .a. positus est rectus. erit per secundam partem et per tertiam .29. primi: uterque angulorum qui sunt ad .d. rectus: ducatur igitur linea .f.a. eritque per quartam primi: linea .a.f. aequalis lineae .b.f. comparatis adinvicem lateribus et angulis triangulorum .a.d.f.b.d.f. et quia linea .b.f. est aequalis lineae .c.f. erunt .3. lineae .b.f.a.f.c.f. adinvicem aequales. quare per .9. tertii erit .f. centrum circuli quaesiti.
¶ Now suppose the triangle a.b.c is amblygonal and angle a obtuse. I divide the side respecting this obtuse angle, b.c, by equals at point h, from where I draw the lines h.d and h.e to the middle points of the two remaining sides, which are d and e, and d.h will be equidistant to a.c and e.h equidistant to a.b according to that which is previously demonstrated in the 39th of the first, namely, that a line dividing two sides of any triangle by equals is equidistant to the third, whereby through the second part of the 29th of the first, each of the two angles, b.d.h and c.e.h, will equal angle a and so each is obtuse, therefore, in drawing the perpendiculars d.f to the line a.b and e.f to the line e.c until they meet at point f, I say that to be the center of the circle.	¶ Sit rursus trigonus .a.b.c. ambligonius. sitque angulus .a. obtusus latus .b.c. respiciens hunc angulorum obtusum. divido per aequalia in puncto .h. a quo ad media puncta duorum reliquorum laterum quae sunt .d. et .e. duco lineas .h.d. et .h.e. eritque .d.h. aequidistans .a.c. et .e.h. aequidistans .a.b. propter id quod demonstratum est supra .39. primi: videlicet quod linea secans duo latera alicuius trianguli per aequalia. tertio est aequidistans: quare per secundam partem .29. primi erit uterque duorum angulorum .b.d.h. et .c.e.h. aequalis angulo .a. et ideo uterque obtusus. ductis igitur perpendicularibus .d.f. ad lineam .a.b. et .e.f. ad lineam .a.c. quousque concurrant in puncto .f. quem dico esse centrum circuli.
For it is manifest they are to meet for the reason previously stated, and the line b.c, which respects the obtuse angle, will cut both of them and they will meet outside the triangle a.b.c. Thus from point f, which is the convergence point of them, I produce the lines f.a, f.b, and f.c that by the fourth of the first will be twice assumed equal: first when the sides and angles of the two triangles a.d.f and b.d.f are compared, and then those of the other two, a.e.f and c.e.f, wherefore by the 9th of the third f is the center of the desired circle.	Manifestum est enim eas concurrere propter causam prius dictam. secabit utraque earum lineam .b.c. quae respicit obtusum et concurrent extra triangulum .a.b.c. igitur a puncto .f. qui est punctus concursus earum: produco lineas .f.a.f.b.f.c. quae per quartam primi bis asumptam erunt aequales comparatis primo lateribus et angulis duorum triangulorum .a.d.f.b.d.f. deinde aliorum duorum .a.e.f.c.e.f. quare per .9. tertii .f. est centrum circuli quaesiti.
¶ Again, suppose that the triangle a.b.c is oxygonal. With all its sides divided by equals, namely, the side a.b at point d, and the side a.c at point e, and b.c at point h, I draw out lines d.e, d.h, and e.h. And d.h is equidistant to a.c and e.h to a.b for that by the previous demonstration in the 39th of the first, and so by the second part of the 39th of the first, both of the angles, b.d.h and c.e.h, will equal the angle a and thus are acute. Therefore, by drawing the perpendiculars, d.f to line a.b and e.f to line a.c, it is plain they shall meet within the triangle a.b.c. And point f shall be the convergence point that I say to be the center of the circle, so I produce the lines f.a, f.b, and f.c, which by the fourth of the first shall as before be twice assumed equal, and so by the 9th of the third f will be the center of the desired circle.	¶ Esto iterum ut trigonus .a.b.c. sit oxigonius divisis omnibus eius lateribus per aequalia: videlicet latus .a.b. in puncto .d. et latus .a.c. in puncto .e. et .b.c. in puncto .h. protraho lineas .d.e.d.h. et .e.h. eritque .d.h. aequidistans .a.c. et .e.h.a.b. propter id quod demonstratum est super .39. primi: quare per secundum partem .39. primi: uterque angulorum .b.d.h.c.e.h. erit aequalis angulo .a. et ideo acutus: ductis igitur perpendicularibus .d.f. ad lineam .a.b. et .e.f. ad lineam .a.c. manifestum est eas concurrere intra triangulum .a.b.c. sitque punctus concursus .f. quem dico esse centrum circuli: produco enim lineas .f.a.f.b.f.c. quae per quartam primi: bis assumptam ut prius erunt aequales: quare per .9. tertii erit .f. centrum circuli quaeseti.
¶ Through the preceding it is evident that if a triangle is orthogonal, the center of the circumscribed circle shall fall in the middle of the side that is opposite the right angle, if amblygonal, the center shall fall without the triangle, and if oxygonal it shall fall within the triangle.	¶ Per praedicta patet quod si triangulus fuerit orthogonius centrum circuli circumscribendi cadet in medio lateris quod opponitur angulo recto. Si fuerit ambloginius centrum cadet extra triangulum. Si autem fuerit oxigonius cadet intra triangulum.
Proposition 6	Propositio .6.
[English] Latin	English [Latin][¶]
To describe a quadrate within a given circle.	Intra datum circulum quadratum describere.
¶ Suppose the given circle is a.b.c.d, whose center is e, within which I want to describe a quadrate. I draw within that two diameters, a.c and b.d, and they cut one another orthogonally over the center e, the ends of which I join by extending the lines a.b, b.c, c.d, and d.a, which I say is to contain the desired quadrate. For the quarter of one will thrice be assumed equal to each of the others on account that the four lines, e.a, e.b, e.c, and e.d are equal, and the four angles that are at e are right, and if each one of the four angles, a, b, c, and d, are right, then by the first part of the 30th of the third, for that each of them is in a semicircle a.b.c.d will be a quadrate by definition, which is the intention.	¶ Sit datus circulus .a.b.c.d. cuius centrum .e. volo intra ipsum describere quadratum. protraho in ipso duas diametros .a.c. et .b.d. secantes se orthogonaliter supra centrum .e. quarum extremitates coniungo protractis lineis .a.b.b.c.c.d. et .d.a. quas dico continere quadratum quaesitum: ipsae enim erunt aequales adinvicem. per quartam primi ter assumptam propter id quod quatuor lineae .e.a.e.b.e.c. et .e.d. sunt aequales. et quatuor anguli qui sunt .a.d.e. [sic] recti. si unusquisque. quatuor angulorum .a.b.c. et .d. est rectus per primam partem .30. tertii: propter id quod quilibet eorum est in semicirculo erit igitur .a.b.c.d. quadratum per diffinitionem quod est propositum.
Proposition 7	Propositio .7.
[English] Latin	English [Latin][¶]
To describe a quadrate around a proposed circle.	Circa propositum circulum quadratum describere.
¶ Suppose the proposed circle is a.b.c.d, whose center is e, around which I want to describe a quadrate. I extend in it two diameters, a.c and b.d, that cut one another orthogonally over the center e, from the extremities of which I draw orthogonal lines at both ends until each of them shall at two sides meet, and the convergence points of them will be f, g, h, and k. And through a corollary of the 15th of the third both of the angles that are at each four of the points, a, b, c, and d, will be right, and because three angles in the quadrilateral a.f.b.e, a, b and e, are right, then the fourth angle, which is f, will be right, since every quadrilateral has four angles equal to four rights as is previously demonstrated by the 32nd of the first. By the same rationale each of the angles, g, h and k, will be right, therefore by the second part of the 28th of the first, the two lines f.g and k.h are equidistant.	¶ Sit propositus circulus .a.b.c.d. cuius centorum .e. volo circa ipsum describere quadratum: protraho in ipso duas diametros .a.c. et .b.d. secantes se orthogonaliter super centrum .e. a quorum extremitatibus duco in utramque partem lineas orthogonaliter quousque quaelibet eaorum concurrant cum duabus lateribus sintque puncta concursus earum .f.g.h.k. eritque per correlarium .15. tertii uterque angulorum qui sunt ad unumquemque quatuor punctorum .a.b.c.d. rectus: quia ergo in quadrilatero .a.f.b.e tres anguli .a.b. et .e. sunt recti: erit quartus angulus qui est .f. rectus: habet enim quodlibet quadrilaterum quatuor. angulos aequales quatuor rectis: ut demonstratum est supra .32. primi: eadem ratione quilibet angulorum .g.h. et .k. erit rectus: ergo per secundam partem .28. primi. duae lineae .f.g. et .k.h.
And likewise are the two f.k and g.h, thus, by the 34th of the first f.k is equal to g.h and f.g to k.h, and that by the same f.k is equal to b.d and f.g to a.c, then indeed, b.d is equal to a.c and the four lines, f.k, g.h, f.g, and k.h, will be equal. But also, the four angles, f, g, k, and h, are right, as was proved formerly, thus f.g.k.h is a quadrate by definition, which is the intention.	Itemque duae .f.k. et .g.h. sunt aequidistantes. ergo per .34. primi .f.k. est aequalis .g.h. et .f.g.k.h. et quia per eandem .f.k. est aequalis .b.d. et .f.g.a.c. At vero b.d. est aequalis .a.c. erunt quatuor lineae .f.k.g.h.f.g. et .k.h. aequales: sed et quatuor anguli .f.g.k.h. sunt recti: ut probatum est prius. ergo .f.g.k.h. est quadratum per diffinitionem quod est propositum.
Proposition 8	Propositio .8.
[English] Latin	English [Latin][¶]
To describe a circle within an assigned quadrate.	Intra quadratum assignatum circulum describere.
¶ Suppose the assigned quadrate is a.b.c.d, within which I want to describe a circle. This is as though a conversion of the 6th. I divide each one its sides by equal, that is, a.d at point f, b.a at point g, c.b at point h, and d.c at point e, and I produce the lines e.g and f.h, cutting one another at point k, which I say to be the center of the circle, for f.h will be equidistant and equal to a.b by the 33nd of the first, on account that a.f and b.h are equal and equidistant.	¶ Sit quadratum assignatum .a.b.c.d. volo intra ipsum describere circulum haec est quasi conversa .6. divido unumquodque latus eius per aequalia .a.d. quidem in puncto .f.b.a. in puncto .g.c.b. in puncto .h. et .d.c. in puncto .e. et produco lineas .e.g. et .f.h. secantes se in puncto .k. quem dico esse centrum circuli. erit enim .f.h. aequidistans et aequalis .a.b per .33. primi: propter id quod .a.f et .d.b.[sic] sunt aequales et aequidistantes.
And similarly by the same, d.c and a.b, and because all four midpoints of those four sides are mutually equal, by the 34th of the first the four lines k.e, k.f, k.g, and k.h will be equal, and so by the 9th of the third k is the center of the desired circle.	Similiter per eandem et .d.c.a.b. et quia omnes medietates quatuor laterum ipsius quadrati sunt adinvicem aequales erunt per .34. primi: quatuor lineae .k.e.k.f.k.g. et .k.h. aequales. ergo per .9. tertii .k. est centrum circuli queasiti.
Proposition 9	Propositio .9.
[English] Latin	English [Latin][¶]
To describe a circle around an assigned quadrate.	Circa assignatum quadratum circulum describere.
¶ Suppose the quadrate is a.b.c.d, around which I want to describe a circle. This is as though a conversion of the 7th.	¶ Sit quadratum .a.b.c.d. volo circa ipsum circulum describere. haec est quasi conversa .7.
I extend in it two diameters, a.c and b.d, cutting one another at point e, which I say to be the center of the circle.	Protraho in ipso duas diametros .a.c. et .b.d. secantes se in puncto .e. quem dico esse centrum circuli.
For when the lines a.d and a.b are equal, by the 5th of the first the angles a.d.b and a.b.d will be equal, and because the whole angle a is right, by the 32nd of the first each of them will be half right.	Cum enim lineae .a.d. et .a.b. sint aequales erunt per .5. primi: anguli .a.d.b. et .a.b.d. aequales. et quia angulus .a. totalis est rectus. erit per .32. primi: uterque eorum medietas recti.
¶ In a similar way it is demonstrated any partial angles contained by the aformentioned diameters and sides of the proposed quadrate to be half right, and so because the angle e.a.d is equal to angle e.d.a, by the 9th of the first the line e.a will equal the line e.d. By the same rationale e.a will equal e.b and e.c will equal e.d, thus because the four lines, e.a, e.b, e.c, and c.d, are equal, by the 9th of the third e will be the center of the desired circle, which is the intention.	¶ Simili quoque modo probabitur quemlibet partialium angulorum a praedictis diametris et lateribus quadrati propositi contentorum esse medietatem recti quia igitur angulus .e.a.d. est aequalis angulo .e.d.a. erit per .9. primi: linea .e.a. aequalis lineae .e.d. eadem ratione erit .e.a. aequalis .e.b. et .e.c. aequalis .e.d. quare quia quatuor lineae .e.a.e.b.e.c.c.d. sunt aequales. erit per .9. tertii e. centrum circuli quaesiti. quod est propositum.
Proposition 10	Propositio .10.
[English] Latin	English [Latin][¶]
To designate a triangle of two equal sides, each of whose two angles, which the base obtains, are twice as much the remaining.	Duum aequalium laterum triangulum designare. cuius uterque duorum angulorum quos basis optinet. reliquo duplus existat.
¶ The intention is to describe one triangle of two equal sides and an unequal third, each of whose angles, which are upon the side that is unequal to the others, are twice as much the third.	¶ Intentio est describere unum triangulum duum aequalium laterum et tertii inaequalis cuius uterque angulorum qui super latus quod est reliquis inaequale existunt ad tertium duplus existat.
And to do this, let any line that is a.b be taken, which is divided at point c according to what the 11th of the second demonstrates, so that that which is produced from a.b according to b.c shall equal the quadrate a.c. With the center made at point a, the circle b.d.e is described following its extent, within which, through the first of this, line b.d is created equal to line a.c and the two lines, d.a and d.c, are produced. I say triangle a.b.d to be such as is proposed. Let the circle that is d.c.a be circumscribed about the triangle d.c.a through the 5th of this, and so that line d.b is equal to line a.c, that produced from a.b according to b.c will equal the quadrate of the line b.d, thus by the last of the third, line b.d is contacting the circle d.c.a, and by the 31st of the same, angle c.d.b is equal to angle c.a.d. Therefore, with c.d.a posited as the common angle, the entire angle b.d.a will equal the two angles c.a.d and c.d.a, but by the 32nd of the first, angle b.c.d is equal to the same, since it is extrinsic to them, thus angle b.d.a is equal to angle b.c.d and since angle a.d.b is equal to angle a.b.d, by the 5th of the first, for that the side a.d and a.b are equal, angle b.c.d will equal angle c.b.d, and so by the 6th of the first the line c.d is equal to the line b.d, and to the line c.a, and so by the 5th of the first the angle c.a.d is equal to the angle c.d.a. Therefore, because both of the angles, c.d.b and c.d.a, are equal to angle c.a.d, the entire angle b.d.a will be twice as much the angle d.a.b, and so angle a.b.d, equal to that, is also twice as much to angle b.a.d, which is the intention.	Ad hoc autem faciendum sumatur. linea quaelibet quae sit .a.b. quae dividatur secundum quod docet .ii. secundi in puncto .c. ita quod illud quod fit ex .a.b. in .b.c. sit aequale quadrato .a.c. facto quod puncto .a. centro secundum ipsius quantitatem describatur circulus .b.d.e. intra quem per primam huius coaptetur linea .b.d. aequalis lineae .a.c. et producantur duae lineae .d.a.d.c. dico triangulum .a.b.d. esse qualis proponitur: circumscribatur circulus qui sit .d.c.a. per .5. huius triangulo .d.c.a. quia ergo linea .d.b. est aequalis lineae .a.c. erit quod fit ex .a.b. in .b.c. aequale quadrato lineae .b.d. quare per ultimam tertii .b.d. linea est contingens circulum .d.c.a. et per .31. eiusdem angulus .c.d.b. est aequalis angulo .c.a.d. posito ergo communi angulo .c.d.a. erit totus angulus .b.d.a. aequalis duobus angulis .c.a.d.c.d.a. sed per .32. primi angulus .b.c.d. est aequalis eisdem quia extrinsecus ad ipsos. ergo angulus .b.d.a. est aequalis angulo .b.c.d. et quia angulus .a.d.b. est aequalis angulo .a.b.d. per .5. primi: eo quod latera .a.d. et .a.b. sunt aequalia. erit angulus .b.c.d. aequalis angulo .c.b.d. ergo per .6. primi: linea .c.d. est aequalis lineae .b.d. quare et lineae .c.a. ergo per .5. primi: angulus .c.a.d. est aequalis angulo .c.d.a. quia ergo uterque angulorum .c.d.b. et .c.d.a. est aequalis angulo .c.a.d. erit totus angulus .b.d.a. duplus ad angulum .d.a.b. et ideo angulus .a.b.d. sibi aequalis. duplus est etiam ad angulum .b.a.d. quod est propositum.
¶ Mayhap the adversary says the circle d.c.a, circumscribed by a partial triangle, cuts the circle b.d.e at any point of the arc b.d so that it simultaneously will cut the line b.d, whereby that will not be applied to the circle as is supposed in the demonstration, but is cutting it.	¶ Forsan dicet adversarius circulum .d.c.a. circumscriptum trigono partiali secare circulum .b.d.e. in aliquo puncto arcus .b.d. ita quod simul secabit lineam .b.d. unde ipsa non erit circulo applicata. sicut in demonstratione supponitur. sed ipsum secans.
Suppose if it were possible as the adversary posits and from point b, b.f is drawn contacting that smaller circle, and the lines f.a and f.d are drawn, then by the penultimate of the third, that which is produced from a.b according to b.c will equal the quadrate b.f, thus b.f is equal to b.d, and so by the 5th of the first angle b.f.d is equal to angle b.d.f. And that by the 31st of third angle b.f.a is equal to angle a.d.f, angle b.d.f will then be greater than angle a.d.f, which is impossible, since it is a part of it.	Sit ergo si possibile est ut ponit adversarius et a puncto .b. ducatur ad ipsum circulum minorem contingens .b.f. et ducantur lineae .f.a.f.d. eritque per penultimam tertii quod fit ex .a.b. in .b.c. aequale quadrato .b.f. ergo .b.f. est aequalis .b.d. quare per .5. primi angulus .b.f.d. est aequalis angulo .b.d.f. et quia per .31. tertii angulus .b.f.a. est aequalis angulo .a.d.f. erit angulus .b.d.f. maior angulo .a.d.f. quod est impossibile. cum ipse sit pars eius.
¶ We may refute this in another way and show that the smaller circle will in no way cut the line b.d.	¶ Aliter possumus istud refellere et ostendere quod ille minor circulus nullo modo secabit lineam .b.d.
So if it is possible that it cuts it, let this happen at point b, that that produced from a.b according to b.c is equal to that produced from d.b according to b.h.	Si enim possibile est quod secet eam. sit hoc in puncto .b. erit quod fit ex .a.b. in .b.c. aequale ei quod fit ex .d.b. in .b.h.
For it is formerly demonstrated by the penultimate of the third that if from any point without the assigned circle any cutting lines are drawn to the circle, then that contained beneath the wholes and their extrinsic portions are mutually equal, and since that made from a.b according to b.c is equal to the quadrate b.d, that made from d.b according to b.h will equal the quadrate d.b, which is impossible by the second of the second, and so the point stands.	Monstratum est enim supra penultimam tertii quod si ab aliquo puncto extra circulum signato quotlibet lineae secantes ad circulum ducantur quae sub totis et earum portionibus extrinsecis continentur. aequalia sunt adinvicem: et quia quod fit ex .a.b. in .b.c. est aequale quadrato .b.d. erit quod fit ex .d.b. in .b.h. aequale quadrato .d.b. quod est impossibile per secundam secundi: quare constat propositum.
¶ Also note that the smaller circle will necessarily cut the greater and will abscind from it one arc equal to the arc b.d, and the greater similarly will abscind from the same one arc equal to the arc d.c.	¶ Et nota quod minor circulus necessario secabit maiorem et abscindet ab eo arcum unum aequalem arcui .b.d. et maior abscindet similiter ab eodem unum arcum aequalem arcui .d.c.
Which is thus proved, for if the smaller does not cut the greater then it contacts it at point d, and that by the 11th of the third the centers of the circles contacting one another and the point of contact are on one line, then the center of the smaller circle will be on the line a.d, on account that on it is the center of the greater and the point of contact, therefore, by the 17th of the third, angle a.d.b is right, whereby angle a.b.d, equal to it, is also right, which is impossible by the 32nd of the first.	Quod sic probatur. si enim minor non secat maiorem. contingit ergo ipsum in puncto .d. et quia per .11. tertii circulorum se contingentium centra. et punctus contactus sunt in linea una. erit centrum minoris circuli in linea .a.d. propter hoc quod in ea est centrum maioris et punctus contactus. ergo per .17. tertii angulus .a.d.b. est rectus quare similiter et angulus .a.b.d. sibi aequalis est rectus quod est impossibile. per .32. primi:
Consequently, it cuts it at points e and d. I say the arc e.d of the greater to be equal to the arc d.b, and the arc e.d of the smaller to be equal to the arc d.c. I produce lines d.e, c.e, and e.a, and by the 26th of the third each of the four angles that are d.e.c, c.e.a, d.a.c, and a.d.c, are equal to one another on account that the two arcs d.c and c.a are equal, by the 27th of the same, whereby the whole angle a.e.d is twice the angle b.a.d, thus equal to both of the angles a.b.d and a.d.b. And because angle a.e.d is equal to angle a.d.e, by the 5th of the first, on account that a.e and a.d are equal from center to circumference, the two angles e and d of triangle a.e.d equal the two angles d and b of triangle a.d.b, so by the 32nd of the first, the remaining angle a of one is equal to the remaining angle a of the other, therefore, by the 25th of the third, the arc e.d of the greater is equal to the arc d.b, and by the same, the arc e.d of the smaller is equal to the arc d.c, and this is what we proposed.	Secet ergo ipsum in punctis .e.d. dico arcum .e.d. maioris esse aequalem arcui .d.b. et arcum .e.d. mionoris esse aequalem arcui .d.c. produco lineas .d.e.c.e. et .e.a. eritque per .26. tertii unusquisque quatuor angulorum qui sunt .d.e.c.c.e.a.d.a.c. et .a.d.c. aequalis alii propter id quod duo arcus .d.e.[sic] et .e.a.[sic] sunt aequales. per .27. eiusdem quare totalis angulus .a.e.d. duplus est ad angulum .b.a.d. et ideo aequalis utrique angulorum .a.b.d. et a.d.b. et quia angulus .a.e.d. est aequalis angulo .a.d.e. per .5. primi: propter id quod .a.e. et .a.d. sunt aequales a centro ad circumferentiam. erunt duo anguli .e. et .d. trianguli .a.e.d. aequales duobus angulis .d. et .b. trianguli .a.d.b. ergo per .32. primi: reliquus angulus .a. unius est aequalis reliquo angulo .a. alterius: ergo per .25. tertii arcus .e.d. maioris est aequalis arcui .d.b. et per eandem arcus .e.d. minoris est aequalis arcui .d.c. et hoc est quod proposuimus.
Proposition 11	Propositio .11.
[English] Latin	English [Latin][¶]
To describe an equilateral and equiangular pentagon within a given circle.	Intra datum circulum aequilaterum. atque aequilangulum penthagonum describere.
¶ Suppose the given circle is a.b.c, within which I want to describe one equilateral and equiangular pentagon. I design one triangle such as is proposed by the preceding, which is 2, and draw another equiangular to it within the given circle, just as the second of this demonstrates, which is a.b.c, and each of the two angles, a.b.c and a.c.b, are twice as much the angle c.a.b, both of which I divide by equals by drawing lines b.e and c.d. And by the 25th of the third the five arcs into which the five points, a, d, b, c and e divide the circle, are mutually equal on account that the five angles that fall on those said arcs are mutually equal, and so by connecting those five points by the rights lines, which are a.d, d.b, b.c, c.e, and e.a, the pentagon a.d.b.c.e will be inscribed within the given circle as proposed, for it is equilateral by the 28th of the third because the 5 arcs, the five sides of which are cords, shall be mutually equal, and it is equiangular as well by the 26th of the same for that the five arcs, d.a.e, a.e.c, e.c.b, c.b.d, and b.d.a, on which the angles of the that pentagon fall, are mutually equal, and thus is the intention.	¶ Sit datus circulus .a.b.c. volo intra ipsum describere penthagonum unum aequilaterum atque aequiangulum. designo triangulum unum qualem praemissa proponit. qui sit .2. cui alium aequiangulum intra datum circulum describo. sicut docet secunda huius: qui sit .a.b.c. sitque uterque angulorum .a.b.c. et .a.c.b. duplus ad angulum .c.a.b. utrumque eorum divido per aequalia ductis lineis .b.e. et .c.d. eruntque per .25. tertii .5. arcus in quos .5. puncta .a.d.b.c.e. dividunt circulum adinvicem aequales. propter id quod quinque anguli qui in dictos arcus cadunt sunt adinvicem aequales. continuatis igitur illis quinque punctis per lineas rectas quae sunt .a.d.d.b.b.c.c.e. et .e.a. erit penthagonus .a.d.b.c.e. inscriptus dato circulo qualis proponitur: est enim aequalaterus per .28. tertii cum .5. arcus: quorum eius quinque latera sunt cordae: sint ad invicem aequales: et etiam aequiangulus per .26. eiusdem eo quod quinque arcus .d.a.e.a.e.c.e.c.b.c.b.d. et .b.d.a. in quos anguli ipsius penthagoni cadunt sunt adinvicem aequales: sicque constat propositum.
Proposition 12	Propositio .12.
[English] Latin	English [Latin][¶]
To designate an equilateral and equiangular pentagon around a proposed circle.	Circa propositum circulum penthagonum aequilaterum atque aequiangulum designare.
¶ Suppose the proposed circle is a.b.c, whose center is f, around which I want to designate an equilateral and equiangular pentagon. Over the circumference of that circle I will note five corner points, as though I had inscribed in it a pentagon according to the doctrine of the preceding, which are a, d, b, c, and e, from the center to which I draw lines f.a, f.d, f.b, f.c, and f.e, and from these same points I shall draw out perpendiculars to those lines in each direction until they meet at points g, h, k, l and m, and these lines are contacting the circle through a corollary of the 15th of the third. And from the center to those convergence points I draw lines f.g, f.h, f.k, f.l, and f.m.	¶ Sit propositus circulus .a.b.c. cuius centrum .f. volo circa ipsum designare penthagonum aequilaterum atque aequiangulum. supra circumferentiam ipsius circuli quasi secundum doctrinam praemissae sibi inscripsissem penthagonum quinque puncta angularia notabo. quae sunt .a.d.b.c.e. ad quae centra ducam lineas .f.a.f.d.f.b.f.c.f.e. et ab eisdem punctis educam perpendiculares ad istas lineas in utranque partem quousque concurrant in punctis .g.h.k.l.m. eruntque hae lineae contingentes circulum per correlarium .15. tertii: et ad ista puncta concursus ducam a centro lineas .f.g.f.h.f.k.f.l.f.m.
And since it is previously shown by the penultimate of the third that if from any point without a circle two lines shall be drawn contacting that circle, because those lines will be equal, the line g.a will equal the line g.d, and h.d h.b, and so on for the others.	Et quia monstratum est super penultimam tertii quod si ab aliquo puncto extra circulum signato duae lineae contingentes ad ipsum circulum ducantur quod ipse erunt aequales. erit linea .g.a. aequalis lineae .g.d. et .h.d.h.b. et sic de ceteris.
And because the five arcs into which the five points, a, d, b, c and, e, divide the circle are mutually equal, by the 26th of the third the five angles, a.f.d, d.f.b, b.f.c, c.f.e, and e.f.a, which lie over these arcs at the center f are themselves mutually equal.	At quoniam quinque arcus in quos quinque puncta .a.d.b.c.e. dividunt circulum. sunt adinvicem aequales. erunt per .26. tertii quinque anguli .a.f.d.d.f.b.b.f.c.c.f.e.e.f.a. consistentes super hos arcus in centro .f. sibi invicem aequales.
But the two sides, a.g and f.a on the triangle f.g.a, are equal to the two sides d.g and f.d of the triangle f.g.d, and the side g.f is common, therefore by the 8th of the first, the two angles of them that are at f are mutually equal.	Sunt autem duo latera .a.g. et .f.a. trianguli .f.g.a aequalia duobus lateribus .d.g. et .f.d. trianguli .f.g.d. et latus .g.f. commune. ergo per .8. primi: duo anguli eorum qui sunt .a.d.f.[sic].
Just like the two angles that are at g are mutually equal, and by the same rationale the two angles that are at f in triangles d.f.h and h.f.b.	Itemque duo anguli qui sunt .a.d.g.[sic] sunt adinvicem aequales. eadem rationem duo anguli qui sunt .a.d.f.[sic] in triangulis .d.f.h. et .h.f.b.
Likewise, the two that are at b are mutually equal.	Itemque duo qui sunt .a.d.b.[sic] sunt adinvicem aequales.
And similarly of each the remaining three angles that are b.f.c, c.f.e, and e.f.a, and of each of the three by which k, l, and m are divided by equals, that is, the first by line f.k, the second by line f.l, and the third by line f.m. And that the three angles that are b.f.c, c.f.e, and e.f.a are themselves mutually equal, and the other two that are a.f.d and d.f.b are equal, then half of them, which are the ten angles made at center f, are mutually equal.	Similiter quoque singuli trium reliquorum angulorum qui sunt .b.f.c.c.f.e.e.f.a. et singuli trium. qui sunt .k.l.m. dividantur per aequalia. primi quidem per lineam .f.k. secundi per lineam .f.l. tertii vero per lineam .f.m. et quia hii tres anguli qui sunt .b.f.c.c.f.e.e.f.a. sunt sibi invicem aequales et aliis duobus qui sunt .a.f.d. et .d.f.b. aequales erunt eorum dimidia quae sunt decem anguli facti in centro .f. adinvicem aequales.
Therefore, because the two angles a and f of the triangle g.a.f are equal to the two angles a and f of the triangle m.a.f, and the side a.f is common, then by the 26th of the first the angle g of one equals angle m of the other and the side g.a equals the side a.m. By the same rationale, angle g on the triangle g.f.d will equal angle h on the triangle d.f.h and the side g.d equals the side d.h, and so because g.a is half g.m, and g.d half g.h, and g.a and g.d are equal, then by common knowledge g.m and g.h will be twice their equal.	Quia igitur duo anguli .a. et .f. trianguli .g.a.f. sunt aequales duobus angulis .a. et .f. trianguli .m.a.f. et latus .a.f. commune erit per .26. primi angulus .g. unius aequalis angulo .m. alterius et latus .g.a. aequale lateri .a.m. eadem ratione erit angulus .g. in triangulo .g.f.d. aequalis angulo .h. in triangulo .d.f.h. et latus .g.d. aequale lateri .d.h. quare quia .g.a. est dimidium .g.m. et .g.d. dimidium .g.h. et .g.a. et .g.d. sunt aequalia: erunt per communem scientiam .g.m. et .g.h. eorum dupla aequalia.
Similarly we shall also demonstrate g.m to be equal to m.l, and m.l to l.k, and l.k to k.h, whereby the pentagon g.h.k.l.m is equilateral. And it is equiangular, for since the two angles that are at g are mutually equal and the two that are at m are similarly mutually equal, and part of g should equal part of m, for both have been proved before, then by the same common knowledge the whole of g will equal the whole of m, and by the same rationale you will show equality in the other angles, whereby it is equiangular, and thus is the intention.	Similiter quoque probabimus .g.m. esse aequale .m.l. et .m.l.l.k. et .l.k.k.h. quare penthagonus .g.h.k.l.m. est aequilaterus. sed et aequiangulus: cum enim duo anguli qui sunt ad .g. sunt adinvicem aequales. et duo qui sunt ad .m. similiter adinvicem aequales. et .g. partialis. sit aequalis .m. partiali. utrumque enim probatum est prius. erit per eandem communem scientiam .g. totalis aequalis .m. totali. et eadem ratione probabis aequilitatem in ceteris angulus: quare est aequiangulus. sicque constat propositum.
Proposition 13	Propositio .13.
[English] Latin	English [Latin][¶]
To describe a circle within an assigned equilateral and equiangular pentagon.	Intra aequilaterum atque aequiangulum penthagonum assignatum. circulum describere.
¶ Suppose the equilateral and equiangular pentagon is assigned (since it is not possible of the others, this is necessary) a.b.c.d.e, to which I want to inscribe a circle. This is as though a conversion of the 11th. I divide two of its neighboring angles, which are a and e, by equals, by drawing lines a.f and e.f until they meet at point f within the pentagon, which I say is to be the center of the circle, for they will meet on account that half the total angle a, and likewise the total angle e, are less than a right angle.	¶ Sit assignatus penthagonus aequilaterus atque aequiangulus: quia de aliis non est necessarium hoc est possibile .a.b.c.d.[sic] volo sibi inscribere circulum. haec est quasi conversa .11. duos eius propinquos angulos qui sunt .a. et .e. divido per aequalia ductis lineis .a.f. et .e.f. donec concurrant in puncto .f. intra ipsum penthagonum quem dico esse centrum circuli: concurrent enim propter id quod dimidium totalis anguli .a. et similiter totalis anguli .e. minus est angulo recto.
¶ For if they do not meet within the pentagon then they shall meet either without the pentagon, or upon the side of the pentagon, or on its angle that is opposite both other angles.	¶ Si enim intra penthagonum non concurrent. aut extra ipsum penthagonum aut in latere penthagoni. aut in eius angulo: qui utrique angulorum diversorum opponitur.
First suppose then they meet without at point f and the line b.f is drawn, and the two sides, e.a and a.f of the triangle e.a.f are equal to the sides b.a and a.f of the triangle b.a.f, and angle a of one equals angle a of the other, so by the 4th of the first the base e.f will equal the base f.b, and since the partial angle a is equal to the partial angle e, for that the whole of a and whole of e are equal, then by the 6th of the first f.a will equal f.e, whereby f.a is equal to f.b, and so by the 6th of the first the two angles, the whole of b and the partial a, are equal, and so a part of a is equal or greater to the whole of a, which is impossible.	Concurrant ergo primo extra in puncto .f. et ducatur linea .b.f. et quia duo latera .e.a. et .a.f. trianguli .e.a.f. sunt aequalia duobus lateribus .b.a. et .a.f. trianguli .b.a.f. et angulus .a. unius angulo .a. alterius erit per .4. primi basis .e.f. aequalis basi .f.b. et quia angulus .a. partialis est aequalis angulo .e. partiali. propter id quod .a. totalis .e. totali erit per .6. primi .f.a. aequalis .f.e. quare .f.a. est aequalis .f.b. ergo per .5.[sic] primi duo anguli .b. totalis. et .a. partialis sunt aequales. quare .a. partialis est aequalis ut maior .a. totali quod est impossibile.
Then suppose they meet at point f on the side b.c. And by arguing as through the preceding and in the manner already given, a part of angle a equals the whole of angle a, which is impossible.	Concurrant ergo in puncto .f. super latus .b.c. eritque arguendo per praemissas et praemisso modo angulus .a. partialis aequalis angulo .a. totali quod est impossibile.
And if they should meet at angle c, then by the same and in the same way c.b will equal c.a and so just as before, a part of angle a is equal to the whole of angle a.	Quod si forsan concurrant in angulo .c. erit per eadem et eodem modo c.b. aequalis .c.a. et ideo ad huc ut prius angulus .a. partialis aequalis angulo .a. totali.
For that this is not possible, suppose then the meeting point, which is f, is within the pentagon, whence I draw 5 perpendiculars to its 5 sides that are f.g, f.h, f.k, f.l, and f.m, and I draw lines f.b and f.d to two of its neighboring angles on the other side, which are divided by equals, which are b and d. And since the two angles a and m of the triangle a.f.m are equal to the two angles a and g of the triangle a.f.g, and the side a.f is common, by the 26th of the first f.m will equal f.g. In the same way you will also prove f.l equal to f.m on the selected triangles e.f.m and e.f.l, for again, the two sides a.f and a.b of the triangle a.f.b are equal to the two sides a.f and a.e of the triangle a.f.e, and the angle a of one equals angle a of the other, then by the 4th of the first the partial angle b will equal the partial angle e, and since the whole of b is equal to the whole of e, the whole of e divided by equals will also equal the whole of b divided by equals.	Quod quia hoc esse non potest sit ergo punctus concursus qui est .f. infra[sic] penthagonum a quo duco .5. perpendiculares ad eius .5. latera. quae sint .f.g.f.h.f.k.f.l.f.m. et ad duos eius angulos propinquos altrinsecus angulis per aequalia divisis qui sunt .b. et .d. duco lineas .f.b.f.d. et quia duo anguli .a. et .m. trianguli .a.f.m. sunt aequales duobus angulis .a. et .g. trianguli .a.f.g. et latus .a.f. commune erit per .26. primi .f.m. aequalis .f.g. per eandem quoque probabis .f.l. aequalem .f.m. sumptis duobus triangulis .e.f.m. et .e.f.l. quia iterum duo latera .a.f. et .a.b. trianguli .a.f.b. sunt aequalia duobus lateribus .a.f. et .a.e. trianguli .a.f.e. et angulus .a. unius. angulo .a: alterius erit per .4. primi angulus .b. partialis aequalis angulo .e. partiali. et quia .b. totalis aequalis est .e. totali: et .e. totalis divisus est per aequalia erit etiam .b. totalis divisus per aequalia.
¶ In the same way you will prove the whole of d is divided by equals, through the equality of partial d and partial a on the selected triangles e.a.f and e.d.f, then because the two angles g and b of the triangle g.f.b are equal to the two angles h and b of the triangle h.f.b, and the side f.b is common, by the 26th of the first f.h will equal f.g.	¶ Eodem modo probabis .d. totalem divisum per aequalia propter aequalitatem .d. partialis et .a. partialis sumptis triangulis .e.a.f. et .e.d.f. quia ergo duo anguli .g. et .b. trianguli .g.f.b. sunt aequales duobus angulis .h. et .b. tringuli .h.f.b. et latus .f.b. commune erit per .26. primi .f.h. aequalis .f.g.
In the same way you will prove f.k equal to f.l on selected the triangles l.f.d and k.f.d, and since the 5 lines, f.g, f.h, f.k, f.l, and f.m are equal, f will be the center of the circle by the 9th of the third, which we will draw following the length of one of them, and it will touch every side of the pentagon according to the equality of the lines and not any of them shall cut by the first part of the 15th of the third, and thus is the intention.	Eodem modo probabis .f.k. aequalem .f.l. sumptis triangulis .l.f.d.k.f.d. quoniam igitur .5. lineae .f.g.f.h.f.k.f.l. et .f.m. sunt aequales. erit .f. centrum circuli. per .9. tertii. quem describemus secundum quantitatem unius earum. et tanget omnia latera penthagoni. propter aequalitatem linearum et nullum eorum secabit per primam partem .15. tertii. sicque constat propositum.
Proposition 14	Propositio .14.
[English] Latin	English [Latin][¶]
To describe a circle around a given pentagon, which is equilateral and equiangular besides.	Circa datum penthagonum quod sit aequilaterum. atque aequiangulum circulum describere.
¶ Suppose as before the given pentagon, both equilateral and equiangular (since it is not possible of the others, this is to be necessary) is a.b.c.d.e, around which I want to describe a circle. This is as if the 12th were reversed.	¶ Sit ut prius datus penthagonus aequilaterus atque aequiangulus. quia de aliis non est necessarium hoc esse possibile .a.b.c.d.e. volo circa ipsum describere circulum. haec est quasi conversa .12.
I divide two of its neighboring angles, which are a and e, by equals by drawing lines a.f and f.e until they meet within the pentagon at point f, for they will meet and that shall happen within the pentagon as is demonstrated in the former, and from the point of contact I draw lines to the remaining angles, which shall be f.b, f.c, and f.d. And because the two sides a.f and a.b of the triangle a.f.b are equal to the two sides a.f and a.e of the triangle a.f.e, and the angle a of one equals angle a of the other, by the 4th of the first f.a will equal f.e, and the partial angle b will equal the partial angle e. And because the whole of b is equal to the whole of a and the whole of e is divided by equals, the whole of b will similarly be divided by equals. In this way you may also prove both of the angles c and d are to be divided by equals, and the 5 lines, f.a, f.b, f.c, f.d, and f.e are to be equal, wherefore by the 9th of the third, f will be the center of the circle, thus the intention is evident.	Duos eius propinquos angulos qui sunt .a. et .e. divido per aequalia ductis lineis .a.f. et .f.e. quousque concurrant intra ipsum penthagonum in puncto .f. concurrent enim et intra penthagonum ut probatum est in praemissa. et a puncto concursus duco ad reliquos angulos lineas quae sint .f.b.f.c.f.d. et quia duo latera .a.f. et .a.b. trianguli .a.f.b. sunt aequalia duobus lateribus .a.f. et .a.e. trianguli .a.f.e. et angulus .a. unius angulo .a. alterius erit per .4. primi .f.a. aequalis .f.e. et angulus .b. partialis angulo .e. partiali. et quia .b. totalis est aequalis .a. totali. et .e. totalis divisus est per aequalia. et quia .b. totalis est aequalis .a. totali. et .e. totalis dividus est per aequalia. erit similiter .b. totalis divisus per aequalia. hoc quoque modo probabis utrumque angulorum .c. et .d. divisum esse per aequalia. et .5. lineas .f.a.f.b.f.c.f.d.f.e. esse aequales. quare per .9. tertii .f. erit centrum circuli. sicque patet propositum.

the 1482 print-edition of Euclid’s Elements ft: Campanus of Novara : in English (& in unabbr Latin)

Book I

The most renowned work of The Elements by Euclid of Megara [sic] together with the commentary of the most perspicacious Campanus on the art of geometry happily begins:

Praeclarissimum opus elementorum Euclidis Megarensis [sic] una cum commentis Campani perspicacissimi in artem geometriam incipit feliciter:

Proposition 1

Propositio .1.

Assemble an equilateral triangle over a given right line.

Triangulum aequilaterum supra datam lineam rectam collocare.

Proposition 2

Propositio .2.

From a given point, to draw a right line equal to any proposed right line.

A dato puncto cuilibet lineae rectae propositae aequam rectam lineam ducere.

Proposition 3

Propositio .3.

From two proposed inequal lines, to abscind from the longer of them one equal to the shorter.

Propositis duabus lineis inaequalibus de longiori earum breviori aequalem abscindere.

Proposition 4

Propositio .4.

Proposition 5

Propositio .5.

It is necessary for all triangles of two equal sides wherewith the angles above the base are equal, that if its two equal sides are extended in a straight line, then the two angles beneath the base likewise will be made equal.

Omnes trianguli duum aequalium laterum angulos qui supra basim sunt aequales esse necesse est: quod si eius duo aequalia latera directe protrahantur fient quoque sub basi duo anguli invicem aequales.

Proposition 6

Propositio .6.

If two angles of any triangle are equal, then the two sides respecting those angles will be equal.

Si duo anguli alicuius trianguli aequales fuerint duoque latera angulos illos respicientia aequalia erunt.

Proposition 7

Propositio .7.

If from the two points terminating any line two lines exit concurrently to one point, it is impossible for any other two lines to be drawn from the same points, each one equal to their conterminate, that might concur at any other in the same direction.

Si a duobus punctis aliquam lineam terminantibus duae lineae ad punctum unum concurrentes exierint ab eisdem punctis alias lineas singulas suis conterminalibus aequales quae ad alium concurrant in eandem partem duci est impossibile.

Proposition 8

Propositio .8.

It is necessary of all two triangles by which the two sides of one will equal the two sides of the other and the base of one will equal the base of the other, for the two angles contained by equal sides to be equal.

Omnium duorum triangulorum quorum duo latera unius duobus lateribus alterius fuerint aequalia: basisque unius basi alterius aequalis: duos angulos aequis contentos aequales esse necesse est:

Proposition 9

Propositio .9.

To cut a given angle by equals.

Datum angulum per aequalia secare.

Proposition 10

Propositio .10.

Proposing a right line to divide it by equals.

Proposita recta linea eam per aequalia dividere.

Proposition 11

Propositio .11.

On a given right line, to draw out from a point indicated on it, a perpendicular supported such from both sides by two equal and right angles.

Data linea recta a puncto in ea signato perpendicularem extrahere duobus quidem angulis aequilibus ac rectis utrique subnixam.

Proposition 12

Propositio .12.

From outside a designated point, to draw a perpendicular to a given line of indefinite length.

A puncto extra signato ad datam lineam indefinite quantitatis perpendicularem deducere.

Proposition 13

Propositio .13.

Of every right line standing over a right line, the two angles on both sides are either right or equal to two right angles.

Omnis rectae lineae super rectam lineam stantis duo utrobique anguli aut sunt recti aut duobus rectis aequales.

Proposition 14

Propositio .14.

If two lines exit from a point of one line in two different directions and produce about them right angles or angles equal to two right angles, those two lines are joined directly with one another and are one line.

Si duae lineae a puncto unius lineae in diversas partes exierint duosque circa se angulos rectos aut duobus rectis aequales fecerint: illae duae lineae sibi directae coniunctae sunt et linea una.

Proposition 15

Propositio .15.

Of all two lines intersecting one another: all angles positioned opposite themselves are equal, whence it is manifest that when two right lines intersect one another, the four angles that are made are to be equal to four right angles.

Omnium duarum linearum se invicem secantium: omnes anguli contra se positi sunt aequales: unde manifestum est quod cum duae lineae recte se invicem secant quatuor qui fiunt angulos quatuor rectis esse aequales.

Proposition 16

Propositio .16.

If any of the sides of a triangle are directly extended, the angle produced from without will be greater than either angle opposite itself from within the triangle.

Si quotlibet laterum trianguli directe protrahatur faciet angulum extrinsecum utroque angulo trianguli sibi intrinsecus opposito maiorem.

Proposition 17

Propositio .17.

Any two angles of every triangle are less than two right angles.

Omnis trianguli duo quilibet anguli duobus rectis sunt minores.

Proposition 18

Propositio .18.

The longest side of every triangle is opposite its greatest angle.

Omnis trianguli longius latus maiori angulo oppositum est:

Proposition 19

Propositio .19.

The greatest angle of every triangle is opposite its longest side.

Omnis trianguli maiori angulo longius latus oppositum est.

Proposition 20

Propositio .20.