Proof: By Euclid

For let the number $A$ neither be (one) of the (numbers) doubled from a dyad, nor let it have an odd half.
I say that $A$ is (both) an even-times-even and an even-times-odd (number).

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In fact, (it is) clear that $A$ is an even-times-even (number) [Def. 7.8] .
For it does not have an odd half.
So I say that it is also an even-times-odd (number).
For if we cut $A$ in half, and (then cut) its half in half, and we do this continually, then we will arrive at some odd number which will measure $A$ according to an even number.
For if not, we will arrive at a dyad, and $A$ will be (one) of the (numbers) doubled from a dyad.
The very opposite thing (was) assumed.
Hence, $A$ is an even-times-odd (number) [Def. 7.9] .
And it was also shown (to be) an even-times-even (number).
Thus, $A$ is (both) an even-times-even and an even-times-odd (number).
(Which is) the very thing it was required to show.
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