第5卷命题 3 · 等倍量之倍量仍等倍

Let a first magnitude A be the same multiple of a second B that a third C is of a fourth D, and let equimultiples EF, GH be taken of A, C; I say that EF is the same multiple of B that GH is of D. For, since EF is the same multiple of A that GH is of C, therefore, as many magnitudes as there are in EF equal to A, so many also are there in GH equal to C.

设第一量A是第二量B的倍量，如同第三量C是第四量D的倍量，并取A与C的等倍量EF与GH。

Let EF be divided into the magnitudes EK, KF equal to A, and GH into the magnitudes GL, LH equal to C; then the multitude of the magnitudes EK, KF will be equal to the multitude of the magnitudes GL, LH.

因EF是A的倍量如同GH是C的倍量，故EF中有多少个等于A的量，GH中也有同样多个等于C的量。

And, since A is the same multiple of B that C is of D, while EK is equal to A, and GL to C, therefore EK is the same multiple of B that GL is of D. For the same reason KF is the same multiple of B that LH is of D.

将EF分为等于A的量EK与KF，GH分为等于C的量GL与LH，则EK、KF的个数等于GL、LH的个数。

Since, then, a first magnitude EK is the same multiple of a second B that a third GL is of a fourth D, and a fifth KF is also the same multiple of the second B that a sixth LH is of the fourth D, therefore the sum of the first and fifth, EF, is also the same multiple of the second B that the sum of the third and sixth, GH, is of the fourth D.

因A是B的倍量如同C是D的倍量，且EK等于A，GL等于C，故EK是B的倍量如同GL是D的倍量；同理KF是B的倍量如同LH是D的倍量；由V.2，EF是B的倍量如同GH是D的倍量。