第5卷命题 17 · 合比与分比互推定理

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.

设AB、BE、CD、DF成合比，即AB比BE等于CD比DF。

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.

取AE、EB、CF、FD的等倍量GH、HK、LM、MN，以及EB、FD的任意等倍量KO、NP。

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.

由V.1，GK是AB的倍量等于LM是CD的倍量；由V.2，HO是EB的倍量等于MP是FD的倍量。

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.

根据合比定义，若GK大于HO则LN大于MP，从而GH大于KO时LM大于NP，故AE比EB等于CF比FD。