Proof: By Euclid
(related to Proposition: 6.04: Equiangular Triangles are Similar)
 Let $BC$ be placed straighton to $CE$.
 And since angles $ABC$ and $ACB$ are less than two right angles [Prop. 1.17], and $ACB$ (is) equal to $DEC$, thus $ABC$ and $DEC$ are less than two right angles.
 Thus, $BA$ and $ED$, being produced, will meet [C.N. 5] .
 Let them have been produced, and let them meet at (point) $F$.
 And since angle $DCE$ is equal to $ABC$, $BF$ is parallel to $CD$ [Prop. 1.28].
 Again, since (angle) $ACB$ is equal to $DEC$, $AC$ is parallel to $FE$ [Prop. 1.28].
 Thus, $FACD$ is a parallelogram.
 Thus, $FA$ is equal to $DC$, and $AC$ to $FD$ [Prop. 1.34].
 And since $AC$ has been drawn parallel to one (of the sides) $FE$ of triangle $FBE$, thus as $BA$ is to $AF$, so $BC$ (is) to $CE$ [Prop. 6.2].
 And $AF$ (is) equal to $CD$.
 Thus, as $BA$ (is) to $CD$, so $BC$ (is) to $CE$, and, alternately, as $AB$ (is) to $BC$, so $DC$ (is) to $CE$ [Prop. 5.16].
 Again, since $CD$ is parallel to $BF$, thus as $BC$ (is) to $CE$, so $FD$ (is) to $DE$ [Prop. 6.2].
 And $FD$ (is) equal to $AC$.
 Thus, as $BC$ is to $CE$, so $AC$ (is) to $DE$, and, alternately, as $BC$ (is) to $CA$, so $CE$ (is) to $ED$ [Prop. 6.2].
 Therefore, since it was shown that as $AB$ (is) to $BC$, so $DC$ (is) to $CE$, and as $BC$ (is) to $CA$, so $CE$ (is) to $ED$, thus, via equality, as $BA$ (is) to $AC$, so $CD$ (is) to $DE$ [Prop. 5.22].
 Thus, in equiangular triangles the sides about the equal angles are proportional, and those (sides) subtending equal angles correspond.
 (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"