Εἰ γὰρ μή, ἐκβαλλόμεναι αἱ ΑΒ, ΓΔ συμπεσοῦνται ἤτοι ἐπὶ τὰ Β, Δ μέρη ἢ ἐπὶ τὰ Α, Γ.
ἐκβεβλήσθωσαν καὶ συμπιπτέτωσαν ἐπὶ τὰ Β, Δ μέρη κατὰ τὸ Η.
τριγώνου δὴ τοῦ ΗΕΖ ἡ ἐκτὸς γωνία ἡ ὑπὸ ΑΕΖ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΕΖΗ: ὅπερ ἐστὶν ἀδύνατον:
οὐκ ἄρα αἱ ΑΒ, ΓΔ ἐκβαλλόμεναι συμπεσοῦνται ἐπὶ τὰ Β, Δ μέρη.
ὁμοίως δὴ δειχθήσεται, ὅτι οὐδὲ ἐπὶ τὰ Α, Γ:
αἱ δὲ ἐπὶ μηδέτερα τὰ μέρη συμπίπτουσαι παράλληλοί εἰσιν:
παράλληλος ἄρα ἐστὶν ἡ ΑΒ τῇ ΓΔ.
For, if not, AB, CD when produced will meet either in the direction of B, D or towards A, C.
Let them be produced and meet, in the direction of B, D, at G.
Then, in the triangle GEF, the exterior angle AEF is equal to the interior and opposite angle EFG: which is impossible. [I. 16]
Therefore AB, CD when produced will not meet in the direction of B, D.
Similarly it can be proved that neither will they meet towards A, C.
But straight lines which do not meet in either direction are parallel; [Def. 23]
therefore AB is parallel to CD.
Si enim non, educte recte AB et GD concident vel in partes BD vel in AG.
Educantur et concidant in BD ad punctum I.
Trigoni ergo IEZ exterior angulus AEZ equalis est interiori et opposito EZI. Quod est impossibile.
Non ergo recte AB et GD educte concidunt in partes BD.
Similiter autem ostendetur quoniam neque in partes AG.
In neutras vero partes concidentes equidistantes sunt.
Equidistans ergo est recta AB recte GD.
برهانه انهما ان لم يكونا متوازيين فانهما اذا اخرجا فى احدى الجهتين التقيا
فنخرجهما فى جهة (ب د) فيلتقيان على نقطة (ك) ان امكن ذلك
فتصير ١١٦ زاوية (ا ح ط) الخارجة من مثلث (ح ط ك) اعظم من زاوية (ح ط ك) الداخلة كما بين ببرهان (يو) من (ا) وهذا خلف لان زاوية (ا ح ط) فرضت مساوية لزاوية (ح ط د)
فخط (ا ب) (ج د) ان اخرجا فى الجهتين جميعا لم يلتقيا ولو خرج الى غير نهاية
فهما متوازيان
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