Theory of proportions: Book 5 Proposition 17
Translations
Ἐὰν συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται. Ἔστω συγκείμενα μεγέθη ἀνάλογον τὰ ΑΒ, ΒΕ, ΓΔ, ΔΖ, ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ: λέγω, ὅτι καὶ διαιρεθέντα ἀνάλογον ἔσται, ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΔΖ. Εἰλήφθω γὰρ τῶν μὲν ΑΕ, ΕΒ, ΓΖ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΗΘ, ΘΚ, ΛΜ, ΜΝ, τῶν δὲ ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια τὰ ΚΞ, ΝΠ. Καὶ ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΘΚ τοῦ ΕΒ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΗΚ τοῦ ΑΒ. ἰσάκις δέ ἐστι πολλαπλάσιον τὸ ΗΘ τοῦ ΑΕ καὶ τὸ ΛΜ τοῦ ΓΖ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΜ τοῦ ΓΖ. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΜΝ τοῦ ΖΔ, ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΛΝ τοῦ ΓΔ. ἰσάκις δὲ ἦν πολλαπλάσιον τὸ ΛΜ τοῦ ΓΖ καὶ τὸ ΗΚ τοῦ ΑΒ: ἰσάκις ἄρα ἐστὶ πολλαπλάσιον τὸ ΗΚ τοῦ ΑΒ καὶ τὸ ΛΝ τοῦ ΓΔ. τὰ ΗΚ, ΛΝ ἄρα τῶν ΑΒ, ΓΔ ἰσάκις ἐστὶ πολλαπλάσια. πάλιν, ἐπεὶ ἰσάκις ἐστὶ πολλαπλάσιον τὸ ΘΚ τοῦ ΕΒ καὶ τὸ ΜΝ τοῦ ΖΔ, ἔστι δὲ καὶ τὸ ΚΞ τοῦ ΕΒ ἰσάκις πολλαπλάσιον καὶ τὸ ΝΠ τοῦ ΖΔ, καὶ συντεθὲν τὸ ΘΞ τοῦ ΕΒ ἰσάκις ἐστὶ πολλαπλάσιον καὶ τὸ ΜΠ τοῦ ΖΔ. Καὶ ἐπεί ἐστιν ὡς τὸ ΑΒ πρὸς τὸ ΒΕ, οὕτως τὸ ΓΔ πρὸς τὸ ΔΖ, καὶ εἴληπται τῶν μὲν ΑΒ, ΓΔ ἰσάκις πολλαπλάσια τὰ ΗΚ, ΛΝ, τῶν δὲ ΕΒ, ΖΔ ἰσάκις πολλαπλάσια τὰ ΘΞ, ΜΠ, εἰ ἄρα ὑπερέχει τὸ ΗΚ τοῦ ΘΞ, ὑπερέχει καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ εἰ ἴσον, ἴσον, καὶ εἰ ἔλαττον, ἔλαττον. ὑπερεχέτω δὴ τὸ ΗΚ τοῦ ΘΞ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΘΚ ὑπερέχει ἄρα καὶ τὸ ΗΘ τοῦ ΚΞ. ἀλλὰ εἰ ὑπερεῖχε τὸ ΗΚ τοῦ ΘΞ, ὑπερεῖχε καὶ τὸ ΛΝ τοῦ ΜΠ: ὑπερέχει ἄρα καὶ τὸ ΛΝ τοῦ ΜΠ, καὶ κοινοῦ ἀφαιρεθέντος τοῦ ΜΝ ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ: ὥστε εἰ ὑπερέχει τὸ ΗΘ τοῦ ΚΞ, ὑπερέχει καὶ τὸ ΛΜ τοῦ ΝΠ. ὁμοίως δὴ δείξομεν, ὅτι κἂν ἴσον ᾖ τὸ ΗΘ τῷ ΚΞ, ἴσον ἔσται καὶ τὸ ΛΜ τῷ ΝΠ, κἂν ἔλαττον, ἔλαττον. καί ἐστι τὰ μὲν ΗΘ, ΛΜ τῶν ΑΕ, ΓΖ ἰσάκις πολλαπλάσια, τὰ δὲ ΚΞ, ΝΠ τῶν ΕΒ, ΖΔ ἄλλα, ἃ ἔτυχεν, ἰσάκις πολλαπλάσια: ἔστιν ἄρα ὡς τὸ ΑΕ πρὸς τὸ ΕΒ, οὕτως τὸ ΓΖ πρὸς τὸ ΖΔ. Ἐὰν ἄρα συγκείμενα μεγέθη ἀνάλογον ᾖ, καὶ διαιρεθέντα ἀνάλογον ἔσται: ὅπερ ἔδει δεῖξαι.
If magnitudes be proportional componendo, they will also be proportional separando. Let AB, BE, CD, DF be magnitudes proportional componendo, so that, as AB is to BE, so is CD to DF; I say that they will also be proportional separando, that is, as AE is to EB, so is CF to DF. For of AE, EB, CF, FD let equimultiples GH, HK, LM, MN be taken, and of EB, FD other, chance, equimultiples, KO, NP. Then, since GH is the same multiple of AE that HK is of EB, therefore GH is the same multiple of AE that GK is of AB. [V. 1] But GH is the same multiple of AE that LM is of CF; therefore GK is the same multiple of AB that LM is of CF. Again, since LM is the same multiple of CF that MN is of FD, therefore LM is the same multiple of CF that LN is of CD. [V. 1] But LM was the same multiple of CF that GK is of AB; therefore GK is the same multiple of AB that LN is of CD. Therefore GK, LN are equimultiples of AB, CD. Again, since HK is the same multiple of EB that MN is of FD, and KO is also the same multiple of EB that NP is of FD, therefore the sum HO is also the same multiple of EB that MP is of FD. [V. 2] And, since, as AB is to BE, so is CD to DF, and of AB, CD equimultiples GK, LN have been taken, and of EB, FD equimultiples HO, MP, therefore, if GK is in excess of HO, LN is also in excess of MP, if equal, equal, and if less, less. Let GK be in excess of HO; then, if HK be subtracted from each, GH is also in excess of KO. But we saw that, if GK was in excess of HO, LN was also in excess of MP; therefore LN is also in excess of MP, and, if MN be subtracted from each, LM is also in excess of NP; so that, if GH is in excess of KO, LM is also in excess of NP. Similarly we can prove that, if GH be equal to KO, LM will also be equal to NP, and if less, less. And GH, LM are equimultiples of AE, CF, while KO, NP are other, chance, equimultiples of EB, FD; therefore, as AE is to EB, so is CF to FD.