Proof: By Euclid
(related to Proposition: 6.21: Similarity of Polygons is Transitive)
 For since $A$ is similar to $C$, ($A$) is equiangular to ($C$), and has the sides about the equal angles proportional [Def. 6.1] .
 Again, since $B$ is similar to $C$, ($B$) is equiangular to ($C$), and has the sides about the equal angles proportional [Def. 6.1] .
 Thus, $A$ and $B$ are each equiangular to $C$, and have the sides about the equal angles proportional [hence, $A$ is also equiangular to $B$, and has the sides about the equal angles proportional].
 Thus, $A$ is similar to $B$ [Def. 6.1] .
 (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"