Triangles, parallels, and area: Book 1 Proposition 20
Translations
Παντὸς τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι. Ἔστω γὰρ τρίγωνον τὸ ΑΒΓ: λέγω, ὅτι τοῦ ΑΒΓ τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι, αἱ μὲν ΒΑ, ΑΓ τῆς ΒΓ, αἱ δὲ ΑΒ, ΒΓ τῆς ΑΓ, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. Διήχθω γὰρ ἡ ΒΑ ἐπὶ τὸ Δ σημεῖον, καὶ κείσθω τῇ ΓΑ ἴση ἡ ΑΔ, καὶ ἐπεζεύχθω ἡ ΔΓ. Ἐπεὶ οὖν ἴση ἐστὶν ἡ ΔΑ τῇ ΑΓ, ἴση ἐστὶ καὶ γωνία ἡ ὑπὸ ΑΔΓ τῇ ὑπὸ ΑΓΔ: μείζων ἄρα ἡ ὑπὸ ΒΓΔ τῆς ὑπὸ ΑΔΓ: καὶ ἐπεὶ τρίγωνόν ἐστι τὸ ΔΓΒ μείζονα ἔχον τὴν ὑπὸ ΒΓΔ γωνίαν τῆς ὑπὸ ΒΔΓ, ὑπὸ δὲ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει, ἡ ΔΒ ἄρα τῆς ΒΓ ἐστι μείζων. ἴση δὲ ἡ ΔΑ τῇ ΑΓ: μείζονες ἄρα αἱ ΒΑ, ΑΓ τῆς ΒΓ: ὁμοίως δὴ δείξομεν, ὅτι καὶ αἱ μὲν ΑΒ, ΒΓ τῆς ΓΑ μείζονές εἰσιν, αἱ δὲ ΒΓ, ΓΑ τῆς ΑΒ. Παντὸς ἄρα τριγώνου αἱ δύο πλευραὶ τῆς λοιπῆς μείζονές εἰσι πάντῃ μεταλαμβανόμεναι: ὅπερ ἔδει δεῖξαι.
In any triangle two sides taken together in any manner are greater than the remaining one. For let ABC be a triangle; I say that in the triangle ABC two sides taken together in any manner are greater than the remaining one, namely BA, AC greater than BC, AB, BC greater than AC, BC, CA greater than AB. For let BA be drawn through to the point D, let DA be made equal to CA, and let DC be joined. Then, since DA is equal to AC, the angle ADC is also equal to the angle ACD; [I. 5] therefore the angle BCD is greater than the angle ADC. [C.N. 5] And, since DCB is a triangle having the angle BCD greater than the angle BDC, and the greater angle is subtended by the greater side, [I. 19] therefore DB is greater than BC. But DA is equal to AC; therefore BA, AC are greater than BC. Similarly we can prove that AB, BC are also greater than CA, and BC, CA than AB. Therefore etc.