Proof: By Euclid
(related to Proposition: 6.15: Characterization of Congruent Triangles)
 For let $CA$ be laid down so as to be straighton (with respect) to $AD$.
 Thus, $EA$ is also straighton (with respect) to $AB$ [Prop. 1.14].
 And let $BD$ have been joined.
 Therefore, since triangle $ABC$ is equal to triangle $ADE$, and $BAD$ (is) some other (triangle), thus as triangle $CAB$ is to triangle $BAD$, so triangle $EAD$ (is) to triangle $BAD$ [Prop. 5.7].
 But, as (triangle) $CAB$ (is) to $BAD$, so $CA$ (is) to $AD$, and as (triangle) $EAD$ (is) to $BAD$, so $EA$ (is) to $AB$ [Prop. 6.1].
 And thus, as $CA$ (is) to $AD$, so $EA$ (is) to $AB$.
 Thus, in triangles $ABC$ and $ADE$ the sides about the equal angles (are) reciprocally proportional.
 And so, let the sides of triangles $ABC$ and $ADE$ be reciprocally proportional, and (thus) let $CA$ be to $AD$, as $EA$ (is) to $AB$.
 I say that triangle $ABC$ is equal to triangle $ADE$.
 For, $BD$ again being joined, since as $CA$ is to $AD$, so $EA$ (is) to $AB$, but as $CA$ (is) to $AD$, so triangle $ABC$ (is) to triangle $BAD$, and as $EA$ (is) to $AB$, so triangle $EAD$ (is) to triangle $BAD$ [Prop. 6.1], thus as triangle $ABC$ (is) to triangle $BAD$, so triangle $EAD$ (is) to triangle $BAD$.
 Thus, (triangles) $ABC$ and $EAD$ each have the same ratio to $BAD$.
 Thus, [triangle] $ABC$ is equal to triangle $EAD$ [Prop. 5.9].
 Thus, in equal triangles also having one angle equal to one (angle) the sides about the equal angles (are) reciprocally proportional.
 And those triangles having one angle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal.
 (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"