Ἐὰν διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται. Ἔστω διῃρημένα μεγέθη ἀνάλογον τὰ ΑΕ, ΕΒ, ΓΖ, ΖΔ, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ: λέγω, ὅτι καὶ συντεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΖΔ. Εἰ γὰρ μή ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, ἔσται ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ ἤτοι πρὸς ἔλασσόν τι τοῦ ΔΖ ἢ πρὸς μεῖζον. Ἔστω πρότερον πρὸς ἔλασσον τὸ ΔΗ. καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΗ, συγκείμενα μεγέθη ἀνάλογόν ἐστιν: ὥστε καὶ διαιρεθέντα ἀνάλογον ἔσται. ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΗ πρὸς τὸ ΗΔ. ὑπόκειται δὲ καὶ ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. καὶ ὡς ἄρα τὸ ΓΗ πρὸς τὸ ΗΔ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. μεῖζον δὲ τὸ πρῶτον τὸ ΓΗ τοῦ τρίτου τοῦ ΓΖ: μεῖζον ἄρα καὶ τὸ δεύτερον τὸ ΗΔ τοῦ τετάρτου τοῦ ΖΔ. ἀλλὰ καὶ ἔλαττον: ὅπερ ἐστὶν ἀδύνατον: οὐκ ἄρα ἐστὶν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς ἔλασσον τοῦ ΖΔ. ὁμοίως δὴ δείξομεν, ὅτι οὐδὲ πρὸς μεῖζον: πρὸς αὐτὸ ἄρα. Ἐὰν ἄρα διῃρημένα μεγέθη ἀνάλογον ᾖ, καὶ συντεθέντα ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι.

If magnitudes be proportional separando, they will also be proportional componendo. Let AE, EB, CF, FD be magnitudes proportional separando, so that, as AE is to EB, so is CF to FD; I say that they will also be proportional componendo, that is, as AB is to BE, so is CD to FD. For, if CD be not to DF as AB to BE, then, as AB is to BE, so will CD be either to some magnitude less than DF or to a greater. First, let it be in that ratio to a less magnitude DG. Then, since, as AB is to BE, so is CD to DG, they are magnitudes proportional componendo; so that they will also be proportional separando. [V. 17] Therefore, as AE is to EB, so is CG to GD. But also, by hypothesis, as AE is to EB, so is CF to FD. Therefore also, as CG is to GD, so is CF to FD. [V. 11] But the first CG is greater than the third CF; therefore the second GD is also greater than the fourth FD. [V. 14] But it is also less: which is impossible. Therefore, as AB is to BE, so is not CD to a less magnitude than FD. Similarly we can prove that neither is it in that ratio to a greater; it is therefore in that ratio to FD itself.