elem.5.5
若一个量是另一个量的同倍量，且从其中减去的部分是从另一个量减去的部分的同倍量，则剩余部分与剩余部分之比等于整体与整体之比。
AB 是 CD 的同倍量；AE 是 CF 的同倍量。在 CD 左侧延长出 G 使 EB 为 CG 的同倍量，再由 V.1 推出 GF = CD，故 GC = FD，余量 EB 是 FD 的同倍量。
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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.
设量AB是量CD的同倍量，且从AB减去的部分AE是从CD减去的部分CF的同倍量。
[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.
取EB为CG的同倍量，使得AE是CF的倍量与EB是CG的倍量相同。
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.
由V.1，AE是CF的倍量与AB是GF的倍量相同；又假设AE是CF的倍量与AB是CD的倍量相同，故AB是GF和CD的同倍量，因此GF等于CD。
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.
两边减去CF得GC等于FD；由于AE是CF的倍量与EB是GC的倍量相同，且GC等于DF，故EB是FD的倍量与AB是CD的倍量相同。