For, since A measures B, let it measure it according to C; I say that C is not odd.

For, if possible, let it be so.

Then, since A measures B according to C, therefore A by multiplying C has made B.

Therefore B is made up of odd numbers the multitude of which is odd.

Therefore B is odd: [[IX. 23](elem.9.23)] which is absurd, for by hypothesis it is even.

Therefore C is not odd; therefore C is even.

Thus A measures B an even number of times.

For this reason then it also measures the half of it. Q. E. D.