Proof: By Euclid

For let it not measure $A$.
And since $D$ is prime, $A$ and $D$ are thus prime to one another [Prop. 7.29].
And as many times as $D$ measures $C$, so many units let there be in $E$.
Therefore, since $D$ measures $C$ according to the units $E$, $D$ has thus made $C$ (by) multiplying $E$ [Def. 7.15] .
But, in fact, $A$ has also made $C$ (by) multiplying $B$.
Thus, the (number created) from (multiplying) $D$ and $E$ is equal to the (number created) from (multiplying) $A$ and $B$.
Thus, as $D$ is to $A$, so $B$ (is) to $E$ [Prop. 7.19].
And $D$ and $A$ (are) prime (to one another), and (numbers) prime (to one another are) also the least (of those numbers having the same ratio) [Prop. 7.21], and the least (numbers) measure those (numbers) having the same ratio (as them) an equal number of times, the greater (measuring) the greater, and the lesser the lesser - that is to say, the leading (measuring) the leading, and the following the following [Prop. 7.20].
Thus, $D$ measures $B$.
So, similarly, we can also show that if ($D$) does not measure $B$ then it will measure $A$.
Thus, $D$ measures one of $A$ and $B$.
(Which is) the very thing it was required to show.
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