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## Translations

Παντὸς τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι. Ἔστω τρίγωνον τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάττονές εἰσι πάντῃ μεταλαμβανόμεναι. Ἐκβεβλήσθω γὰρ ἡ ΒΓ ἐπὶ τὸ Δ. Καὶ ἐπεὶ τριγώνου τοῦ ΑΒΓ ἐκτός ἐστι γωνία ἡ ὑπὸ ΑΓΔ, μείζων ἐστὶ τῆς ἐντὸς καὶ ἀπεναντίον τῆς ὑπὸ ΑΒΓ. κοινὴ προσκείσθω ἡ ὑπὸ ΑΓΒ: αἱ ἄρα ὑπὸ ΑΓΔ, ΑΓΒ τῶν ὑπὸ ΑΒΓ, ΒΓΑ μείζονές εἰσιν. ἀλλ' αἱ ὑπὸ ΑΓΔ, ΑΓΒ δύο ὀρθαῖς ἴσαι εἰσίν: αἱ ἄρα ὑπὸ ΑΒΓ, ΒΓΑ δύο ὀρθῶν ἐλάσσονές εἰσιν. ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ ὑπὸ ΒΑΓ, ΑΓΒ δύο ὀρθῶν ἐλάσσονές εἰσι καὶ ἔτι αἱ ὑπὸ ΓΑΒ, ΑΒΓ. Παντὸς ἄρα τριγώνου αἱ δύο γωνίαι δύο ὀρθῶν ἐλάσσονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι.

In any triangle two angles taken together in any manner are less than two right angles. Let ABC be a triangle; I say that two angles of the triangle ABC taken together in any manner are less than two right angles. For let BC be produced to D. [Post. 2] Then, since the angle ACD is an exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC. [I. 16] Let the angle ACB be added to each; therefore the angles ACD, ACB are greater than the angles ABC, BCA. But the angles ACD, ACB are equal to two right angles. [I. 13] Therefore the angles ABC, BCA are less than two right angles. Similarly we can prove that the angles BAC, ACB are also less than two right angles, and so are the angles CAB, ABC as well. Therefore etc.