For, since AB is the same multiple of C that DE is of F, as many magnitudes as there are in AB equal to C, so many are there also in DE equal to F.

Let AB be divided into the magnitudes AG, GH, HB equal to C, and DE into the magnitudes DK, KL, LE equal to F; then the multitude of the magnitudes AG, GH, HB will be equal to the multitude of the magnitudes DK, KL, LE.

And, since AG. GH, HB are equal to one another, and DK, KL, LE are also equal to one another, therefore, as AG is to DK, so is GH to KL, and HB to LE. [[V. 7](elem.5.7)]

Therefore, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents; [[V. 12](elem.5.12)] therefore, as AG is to DK, so is AB to DE.

But AG is equal to C and DK to F; therefore, as C is to F, so is AB to DE.

Therefore etc. Q. E. D.