For since, whatever parts AB is of C, DE is also the same parts of F,
therefore, as many parts of C as there are in AB, so many parts of F are there also in DE.
Let AB be divided into the parts of C, namely AG, GB, and DE into the parts of F, namely DH, HE;
thus the multitude of AG, GB will be equal to the multitude of DH, HE.
And since, whatever part AG is of C, the same part is DH of F also, therefore,
whatever part AG is of C, the same part also is the sum of AG, DH of the sum of C, F. [VII. 5]
For the same reason, whatever part GB is of C, the same part also is the sum of GB, HE of the sum of C, F.
Therefore, whatever parts AB is of C, the same parts also is the sum of AB, DE of the sum of C, F.
Q. E. D.