Book 5 Proposition 14
Ἐὰν πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον. Πρῶτον γὰρ τὸ Α πρὸς δεύτερον τὸ Β τὸν αὐτὸν ἐχέτω λόγον καὶ τρίτον τὸ Γ πρὸς τέταρτον τὸ Δ, μεῖζον δὲ ἔστω τὸ Α τοῦ Γ: λέγω, ὅτι καὶ τὸ Β τοῦ Δ μεῖζόν ἐστιν. Ἐπεὶ γὰρ τὸ Α τοῦ Γ μεῖζόν ἐστιν, ἄλλο δέ, ὃ ἔτυχεν, [μέγεθος] τὸ Β, τὸ Α ἄρα πρὸς τὸ Β μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. ὡς δὲ τὸ Α πρὸς τὸ Β, οὕτως τὸ Γ πρὸς τὸ Δ: καὶ τὸ Γ ἄρα πρὸς τὸ Δ μείζονα λόγον ἔχει ἤπερ τὸ Γ πρὸς τὸ Β. πρὸς ὃ δὲ τὸ αὐτὸ μείζονα λόγον ἔχει, ἐκεῖνο ἔλασσόν ἐστιν: ἔλασσον ἄρα τὸ Δ τοῦ Β: ὥστε μεῖζόν ἐστι τὸ Β τοῦ Δ. Ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ Α τῷ Γ, ἴσον ἔσται καὶ τὸ Β τῷ Δ, κἂν ἔλασσον ᾖ τὸ Α τοῦ Γ, ἔλασσον ἔσται καὶ τὸ Β τοῦ Δ. Ἐὰν ἄρα πρῶτον πρὸς δεύτερον τὸν αὐτὸν ἔχῃ λόγον καὶ τρίτον πρὸς τέταρτον, τὸ δὲ πρῶτον τοῦ τρίτου μεῖζον ᾖ, καὶ τὸ δεύτερον τοῦ τετάρτου μεῖζον ἔσται, κἂν ἴσον, ἴσον, κἂν ἔλαττον, ἔλαττον: ὅπερ ἔδει δεῖξαι.
If a first magnitude have to a second the same ratio as a third has to a fourth, and the first be greater than the third, the second will also be greater than the fourth; if equal, equal; and if less, less. For let a first magnitude A have the same ratio to a second B as a third C has to a fourth D; and let A be greater than C; I say that B is also greater than D. For, since A is greater than C, and B is another, chance, magnitude, therefore A has to B a greater ratio than C has to B. [V. 8] But, as A is to B, so is C to D; therefore C has also to D a greater ratio than C has to B. [V. 13] But that to which the same has a greater ratio is less; [V. 10] therefore D is less than B; so that B is greater than D. Similarly we can prove that, if A be equal to C, B will also be equal to D; and, if A be less than C, B will also be less than D.