Καὶ ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς ἐμπέπτωκεν ἡ ΑΓ, αἱ ἐναλλὰξ γωνίαι αἱ ὑπὸ ΒΑΓ, ΑΓΕ ἴσαι ἀλλήλαις εἰσίν.
πάλιν, ἐπεὶ παράλληλός ἐστιν ἡ ΑΒ τῇ ΓΕ, καὶ εἰς αὐτὰς ἐμπέπτωκεν εὐθεῖα ἡ ΒΔ, ἡ ἐκτὸς γωνία ἡ ὑπὸ ΕΓΔ ἴση ἐστὶ τῇ ἐντὸς καὶ ἀπεναντίον τῇ ὑπὸ ΑΒΓ.
ἐδείχθη δὲ καὶ ἡ ὑπὸ ΑΓΕ τῇ ὑπὸ ΒΑΓ ἴση:
ὅλη ἄρα ἡ ὑπὸ ΑΓΔ γωνία ἴση ἐστὶ δυσὶ ταῖς ἐντὸς καὶ ἀπεναντίον ταῖς ὑπὸ ΒΑΓ, ΑΒΓ.
Then, since AB is parallel to CE, and AC has fallen upon them, the alternate angles BAC, ACE are equal to one another. [I. 29]
Again, since AB is parallel to CE, and the straight line BD has fallen upon them, the exterior angle ECD is equal to the interior and opposite angle ABC. [I. 29]
But the angle ACE was also proved equal to the angle BAC;
therefore the whole angle ACD is equal to the two interior and opposite angles BAC, ABC.
Et quoniam equidistans est recta AB recte GE et in ipsas incidit recta AG, anguli ergo BAG et AGE qui permutatim alternis sunt equales.
Rursum quoniam equidistans est recta AB recte GE et in ipsas incidit recta BD, exterior angulus EGD equalis est interiori et opposito qui est ABG.
Demonstratus autem et angulus AGE angulo BAG equalis.
Totus ergo AGD exterior angulus equalis est duobus qui interius et ex adverso qui sunt BAG et ABG.
فخط (ا ج) مخرج على خطى (ا ب) (ج ه) المتوازيين فببرهان (كط) من (ا) زاويتا (ب ا ج) (ا ج ه) المتبادلتان متساويتان
وايضا فانه قد اخرج خط على خطى (ا ب) (ج ه) المتوازيين فزاويتا (ا ب د) (ه ج د) المتقابلتان متساويتن ببرهان (كط) من (ا)
وقد بينا ان زاوية (ا ج ه) مساوية لزاوية (ب ا ج)
tatra ajahakoṇo baajakoṇena tulyo jātaḥ |
hajadakoṇo bakoṇena tulyo jātaḥ |
bahiḥstaḥ baakoṇadvayayogena tulyo jātaḥ |
Complete text
Book I