Proof: By Euclid
(related to Proposition: Prop. 9.16: Two Coprime Integers have no Third Integer Proportional)
 For, if possible, let it be that as $A$ (is) to $B$, (so) $B$ (is) to $C$.
 And $A$ and $B$ (are) prime (to one another).
 And (numbers) prime (to one another are) also the least (of those numbers having the same ratio as them) [Prop. 7.21].
 And the least numbers measure those (numbers) having the same ratio (as them) an equal number of times, the leading (measuring) the leading, and the following the following [Prop. 7.20].
 Thus, $A$ measures $B$, as the leading (measuring) the leading.
 And ($A$) also measures itself.
 Thus, $A$ measures $A$ and $B$, which are prime to one another.
 The very thing (is) absurd.
 Thus, as $A$ (is) to $B$, so $B$ cannot be to $C$.
 (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"