Triangles, parallels, and area: Book 1 Proposition 19
Translations
Παντὸς τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει. Ἔστω τρίγωνον τὸ ΑΒΓ μείζονα ἔχον τὴν ὑπὸ ΑΒΓ γωνίαν τῆς ὑπὸ ΒΓΑ: λέγω, ὅτι καὶ πλευρὰ ἡ ΑΓ πλευρᾶς τῆς ΑΒ μείζων ἐστίν. Εἰ γὰρ μή, ἤτοι ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ ἢ ἐλάσσων: ἴση μὲν οὖν οὐκ ἔστιν ἡ ΑΓ τῇ ΑΒ: ἴση γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῇ ὑπὸ ΑΓΒ: οὐκ ἔστι δέ: οὐκ ἄρα ἴση ἐστὶν ἡ ΑΓ τῇ ΑΒ. οὐδὲ μὴν ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ: ἐλάσσων γὰρ ἂν ἦν καὶ γωνία ἡ ὑπὸ ΑΒΓ τῆς ὑπὸ ΑΓΒ: οὐκ ἔστι δέ: οὐκ ἄρα ἐλάσσων ἐστὶν ἡ ΑΓ τῆς ΑΒ. ἐδείχθη δέ, ὅτι οὐδὲ ἴση ἐστίν. μείζων ἄρα ἐστὶν ἡ ΑΓ τῆς ΑΒ. Παντὸς ἄρα τριγώνου ὑπὸ τὴν μείζονα γωνίαν ἡ μείζων πλευρὰ ὑποτείνει: ὅπερ ἔδει δεῖξαι.
In any triangle the greater angle is subtended by the greater side. Let ABC be a triangle having the angle ABC greater than the angle BCA; I say that the side AC is also greater than the side AB. For, if not, AC is either equal to AB or less. Now AC is not equal to AB; for then the angle ABC would also have been equal to the angle ACB; [I. 5] but it is not; therefore AC is not equal to AB. Neither is AC less than AB, for then the angle ABC would also have been less than the angle ACB; [I. 18] but it is not; therefore AC is not less than AB. And it was proved that it is not equal either. Therefore AC is greater than AB. Therefore etc.