Proof: By Euclid
(related to Proposition: 6.32: Triangles with Two Sides Parallel and Equal)
 For since $AB$ is parallel to $DC$, and the straight line $AC$ has fallen across them, the alternate angles $BAC$ and $ACD$ are equal to one another [Prop. 1.29].
 So, for the same (reasons), $CDE$ is also equal to $ACD$.
 And, hence, $BAC$ is equal to $CDE$.
 And since $ABC$ and $DCE$ are two triangles having the one angle at $A$ equal to the one angle at $D$, and the sides about the equal angles proportional, (so that) as $BA$ (is) to $AC$, so $CD$ (is) to $DE$, triangle $ABC$ is thus equiangular to triangle $DCE$ [Prop. 6.6].
 Thus, angle $ABC$ is equal to $DCE$.
 And (angle) $ACD$ was also shown (to be) equal to $BAC$.
 Thus, the whole (angle) $ACE$ is equal to the two (angles) $ABC$ and $BAC$.
 Let $ACB$ have been added to both.
 Thus, $ACE$ and $ACB$ are equal to $BAC$, $ACB$, and $CBA$.
 But, $BAC$, $ABC$, and $ACB$ are equal to two right angles [Prop. 1.32].
 Thus, $ACE$ and $ACB$ are also equal to two right angles.
 Thus, the two straight lines $BC$ and $CE$, not lying on the same side, make adjacent angles $ACE$ and $ACB$ (whose sum is) equal to two right angles with some straight line $AC$, at the point $C$ on it.
 Thus, $BC$ is straighton to $CE$ [Prop. 1.14].
 Thus, if two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straighton (with respect to one another).
 (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"