Theory of proportions: Book 5 Proposition 5
Translations
Ἐὰν μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι τὸ ὅλον τοῦ ὅλου. Μέγεθος γὰρ τὸ ΑΒ μεγέθους τοῦ ΓΔ ἰσάκις ἔστω πολλαπλάσιον, ὅπερ ἀφαιρεθὲν τὸ ΑΕ ἀφαιρεθέντος τοῦ ΓΖ: λέγω, ὅτι καὶ λοιπὸν τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. Ὁσαπλάσιον γάρ ἐστι τὸ ΑΕ τοῦ ΓΖ, τοσαυταπλάσιον γεγονέτω καὶ τὸ ΕΒ τοῦ ΓΗ. Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΗΓ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΗΖ. κεῖται δὲ ἰσάκις πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ. ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΒ ἑκατέρου τῶν ΗΖ, ΓΔ: ἴσον ἄρα τὸ ΗΖ τῷ ΓΔ. κοινὸν ἀφῃρήσθω τὸ ΓΖ: λοιπὸν ἄρα τὸ ΗΓ λοιπῷ τῷ ΖΔ ἴσον ἐστίν. καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΗΓ, ἴσον δὲ τὸ ΗΓ τῷ ΔΖ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΕΒ τοῦ ΖΔ. ἰσάκις δὲ ὑπόκειται πολλαπλάσιον τὸ ΑΕ τοῦ ΓΖ καὶ τὸ ΑΒ τοῦ ΓΔ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΕΒ τοῦ ΖΔ καὶ τὸ ΑΒ τοῦ ΓΔ. καὶ λοιπὸν ἄρα τὸ ΕΒ λοιποῦ τοῦ ΖΔ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστιν ὅλον τὸ ΑΒ ὅλου τοῦ ΓΔ. Ἐὰν ἄρα μέγεθος μεγέθους ἰσάκις ᾖ πολλαπλάσιον, ὅπερ ἀφαιρεθὲν ἀφαιρεθέντος, καὶ τὸ λοιπὸν τοῦ λοιποῦ ἰσάκις ἔσται πολλαπλάσιον, ὁσαπλάσιόν ἐστι καὶ τὸ ὅλον τοῦ ὅλου: ὅπερ ἔδει δεῖξαι.
If a magnitude be the same multiple of a magnitude that a part subtracted is of a part subtracted, the remainder will also be the same multiple of the remainder that the whole is of the whole. For let the magnitude AB be the same multiple of the magnitude CD that the part AE subtracted is of the part CF subtracted; I say that the remainder EB is also the same multiple of the remainder FD that the whole AB is of the whole CD. For, whatever multiple AE is of CF, let EB be made that multiple of CG. Then, since AE is the same multiple of CF that EB is of GC, therefore AE is the same multiple of CF that AB is of GF. [V. 1] But, by the assumption, AE is the same multiple of CF that AB is of CD. Therefore AB is the same multiple of each of the magnitudes GF, CD; therefore GF is equal to CD. Let CF be subtracted from each; therefore the remainder GC is equal to the remainder FD. And, since AE is the same multiple of CF that EB is of GC, and GC is equal to DF, therefore AE is the same multiple of CF that EB is of FD. But, by hypothesis, AE is the same multiple of CF that AB is of CD; therefore EB is the same multiple of FD that AB is of CD. That is, the remainder EB will be the same multiple of the remainder FD that the whole AB is of the whole CD.