Proof: By Euclid
(related to Proposition: 7.30: Euclidean Lemma)
 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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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"