For, since AB is the same multiple of E that CD is of F, as many magnitudes as there are in AB equal to E, so many also are there in CD equal to F.
Let AB be divided into the magnitudes AG, GB equal to E, and CD into CH, HD equal to F;
then the multitude of the magnitudes AG, GB will be equal to the multitude of the magnitudes CH, HD.
Now, since AG is equal to E, and CH to F, therefore AG is equal to E, and AG, CH to E, F.
For the same reason GB is equal to E, and GB, HD to E, F;
therefore, as many magnitudes as there are in AB equal to E, so many also are there in AB, CD equal to E, F;
therefore, whatever multiple AB is of E, that multiple will AB, CD also be of E, F.