Proof: By Euclid

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 straight-on 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 straight-on (with respect to one another).
(Which is) the very thing it was required to show.
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