For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.
Therefore the triangle ABC is equal to the triangle FCE;
for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38]
But the triangle ABC is equal to the triangle DCE;
therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible.
Therefore AF is not parallel to BE.
Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.
Si enim non est, cadet parallela ipsi B F, per A, ducta, vel supra A D, vel infra. Cadat primum supra, coeatque cum E D, producta in G, & ducatur recta G F.
Quoniam igitur parallelæ sunt A G, B F, erit triangulum E F G, triangulo A B C, æquale:
Ponitur autem & triangulum D E F, eidem triangulo A B C, æquale.
Igitur triangula D E F, G E F, æqualia erunt, pars & totum. Quod est absurdum.
Quod si parallela ducta per A, cadat infra A D, qualis est A H; ducta recta H F, erunt eadem argumentatione triangula H E F, D E F, æqualia, pars & totum, quod est ab?urdum. Est igitur A D, parallela ipsi B F.
Complete text
Book I