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Triangles which are on the same base and in the same parallels are equal to one another.
| Τὰ τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν. Ἔστω τρίγωνα τὰ ΑΒΓ, ΔΒΓ ἐπὶ τῆς αὐτῆς βάσεως τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΑΔ, ΒΓ: λέγω, ὅτι ἴσον ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. Ἐκβεβλήσθω ἡ ΑΔ ἐφ' ἑκάτερα τὰ μέρη ἐπὶ τὰ Ε, Ζ, καὶ διὰ μὲν τοῦ Β τῇ ΓΑ παράλληλος ἤχθω ἡ ΒΕ, διὰ δὲ τοῦ Γ τῇ ΒΔ παράλληλος ἤχθω ἡ ΓΖ. παραλληλόγραμμον ἄρα ἐστὶν ἑκάτερον τῶν ΕΒΓΑ, ΔΒΓΖ: καί εἰσιν ἴσα: ἐπί τε γὰρ τῆς αὐτῆς βάσεώς εἰσι τῆς ΒΓ καὶ ἐν ταῖς αὐταῖς παραλλήλοις ταῖς ΒΓ, ΕΖ: καί ἐστι τοῦ μὲν ΕΒΓΑ παραλληλογράμμου ἥμισυ τὸ ΑΒΓ τρίγωνον: ἡ γὰρ ΑΒ διάμετρος αὐτὸ δίχα τέμνει: τοῦ δὲ ΔΒΓΖ παραλληλογράμμου ἥμισυ τὸ ΔΒΓ τρίγωνον: ἡ γὰρ ΔΓ διάμετρος αὐτὸ δίχα τέμνει. [ τὰ δὲ τῶν ἴσων ἡμίση ἴσα ἀλλήλοις ἐστίν ]. ἴσον ἄρα ἐστὶ τὸ ΑΒΓ τρίγωνον τῷ ΔΒΓ τριγώνῳ. Τὰ ἄρα τρίγωνα τὰ ἐπὶ τῆς αὐτῆς βάσεως ὄντα καὶ ἐν ταῖς αὐταῖς παραλλήλοις ἴσα ἀλλήλοις ἐστίν: ὅπερ ἔδει δεῖξαι. | Triangles which are on the same base and in the same parallels are equal to one another. Let ABC, DBC be triangles on the same base BC and in the same parallels AD, BC; I say that the triangle ABC is equal to the triangle DBC. Let AD be produced in both directions to E, F; through B let BE be drawn parallel to CA, [I. 31] and through C let CF be drawn parallel to BD. [I. 31] Then each of the figures EBCA, DBCF is a parallelogram; and they are equal, for they are on the same base BC and in the same parallels BC, EF. [I. 35] Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34] And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34] [But the halves of equal things are equal to one another.] Therefore the triangle ABC is equal to the triangle DBC. Therefore etc. |