Triangles, parallels, and area: Book 1 Proposition 38
Translations
Τὰ τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν. Ἔστω τρίγωνα τὰ ΑΒΓ, ΔΕΖ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΑΔ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. Ἐκβεβλήσθω γὰρ ἡ ΑΔ ἐφ' ἑκάτερα τὰ μέρη ἐπὶ τὰ Η, Θ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΗ, διὰ δὲ τοῦ Ζ τῇ ΔΕ παράλληλος ἤχθω ἡ ΖΘ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΗΒΓΑ, ΔΕΖΘ: καὶ ἴσον τὸ ΗΒΓΑ τῷ ΔΕΖΘ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΕΖ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΖ, ΗΘ: καί ἐστι τοῦ μὲν ΗΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον. ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει: τοῦ δὲ ΔΕΖΘ παραλληλογράμμου ἥμισυ τὸ ΖΕΔ τρίγωνον: ἡ γὰρ ΔΖ διάμετρος αὐτὸ δίχα τέμνει: [τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν]. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΕΖ τριγώνῳ. Τὰ ἄρα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι.
Triangles which are on equal bases and in the same parallels are equal to one another. Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF. For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE. Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH; for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36] Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34] And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34] [But the halves of equal things are equal to one another.] Therefore the triangle ABC is equal to the triangle DEF. Therefore etc.