Then, since AF is equal to AG and AB to AC, the two sides FA, AC are equal to the two sides GA, AB, respectively;
and they contain a common angle, the angle FAG.
Therefore the base FC is equal to the base GB,
and the triangle AFC is equal to the triangle AGB,
and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend,
that is, the angle ACF to the angle ABG, and the angle AFC to the angle AGB. [I. 4]
And, since the whole AF is equal to the whole AG, and in these AB is equal to AC, the remainder BF is equal to the remainder CG.
But FC was also proved equal to GB;
therefore the two sides BF, FC are equal to the two sides CG, GB respectively;
and the angle BFC is equal to the angle CGB, while the base BC is common to them;
therefore the triangle BFC is also equal to the triangle CGB, and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend;
therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG.
Accordingly, since the whole angle ABG was proved equal to the angle ACF,
and in these the angle CBG is equal to the angle BCF,
the remaining angle ABC is equal to the remaining angle ACB;
and they are at the base of the triangle ABC.
But the angle FBC was also proved equal to the angle GCB;
and they are under the base.
Quia ergo duo latera A B, A F, trianguli A B F, æqualia sunt duobus lateribus A C, A D, trianguli A C D, utrumque utrique, nempe A B, ipsi A C, ex hypothesi, & A F, ipsi A D, ex constructione; angulusque A, contentus lateribus A B, A F, æqualis est angulo A, contento lateribus A C, A D,
immo angulus A, communis est utrique triangulo:
Erit basis B F, æqualis basi C D;
& angulus F, angulo D; & angulus A B F, angulo A C D; cum & priores duo, & posteriores opponantur æqualibus lateribus in dictis triangulis, ut patet. Rursus considerentur duo triangula B D C, C F B.
Quoniam vero rectæ A D, A F, æquales sunt, per constructionem, fit ut, si auferantur ex ipsis æquales A B, A C, & reliquæ B D, & C F, sint æquales.
Quare duo latera B D, D C, trianguli B D C, æqualia sunt duobus lateribus C F, F B, trianguli C F B, utrumque utrique, videlicet B D, ipsi C F, & D C, ipsi F B, ut probatum est.
Sunt autem & anguli D, & F, contenti dictis lateribus æqualibus æquales, ut ostensum etiam fuit. Igitur erit angulus D B C, angulo F C B, æqualis; & angulus B C D, angulo C B F. Tam enim priores duo, quam posteriores, æqualibus opponunturlateribus, existuntque supra communem basim B C, utriusque trianguli B D C, C F B.
Quod si ex totis angulis æqualibus A B F, A C D, (quos æquales esse iam demonstrauimus in prioribus triangulis) detrahantur anguli æquales C B F, B C D, (quos itidem in posterioribus triangulis modo probauimus esse æquales) remanebunt anguli A B C, A C B, supra basim B C, æquales.
Ostensum est autem in posterioribus triangulis, & angulos D B C, F C B, qui quidem sunt infra eandem basim B C, esse æquales.
論曰。試如甲戊線稍長。卽從甲戊截取一分。與甲丁等。為甲己。本篇三 次自丙至丁乙至己。各作直線。第一求卽甲己乙、甲丁丙、兩三角形必等。
何者此兩形之甲角同。
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Complete text
Book I