## Book I, Proposition 40

Equal triangles which are on equal bases and on the same side are also in the same parallels.

Τὰ ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. Ἔστω ἴσα τρίγωνα τὰ ΑΒΓ, ΓΔΕ ἐπὶ ἴσων βάσεων τῶν ΒΓ, ΓΕ καὶ ἐπὶ τὰ αὐτὰ μέρη. λέγω, ὅτι καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν. Ἐπεζεύχθω γὰρ ἡ ΑΔ: λέγω, ὅτι παράλληλός ἐστιν ἡ ΑΔ τῇ ΒΕ. Εἰ γὰρ μή, ἤχθω διὰ τοῦ Α τῇ ΒΕ παράλληλος ἡ ΑΖ, καὶ ἐπεζεύχθω ἡ ΖΕ. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΖΓΕ τριγώνῳ: ἐπί τε γὰρ ἴσων βάσεών εἰσι τῶν ΒΓ, ΓΕ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΕ, ΑΖ. ἀλλὰ τὸ ΑΒΓ τρίγωνον ἴσον ἐστὶ τῷ ΔΓΕ [ τριγώνῳ ]: καὶ τὸ ΔΓΕ ἄρα [ τρίγωνον ] ἴσον ἐστὶ τῷ ΖΓΕ τριγώνῳ τὸ μεῖζον τῷ ἐλάσσονι: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα παράλληλος ἡ ΑΖ τῇ ΒΕ. ὁμοίως δὴ δείξομεν, ὅτι οὐδ' ἄλλη τις πλὴν τῆς ΑΔ: ἡ ΑΔ ἄρα τῇ ΒΕ ἐστι παράλληλος. Τὰ ἄρα ἴσα τρίγωνα τὰ ἐπὶ ἴσων βάσεων ὄντα καὶ ἐπὶ τὰ αὐτὰ μέρη καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι. | Equal triangles which are on equal bases and on the same side are also in the same parallels. Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side. I say that they are also in the same parallels. For let AD be joined; I say that AD is parallel to BE. For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined. Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38] But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible. Therefore AF is not parallel to BE. Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE. |