Proof: By Euclid

Let $BC$ have been joined.
And since $AB$ is parallel to $CD$, and $BC$ has fallen across them, the alternate angles $ABC$ and $BCD$ are equal to one another [Prop. 1.29].
And since $AB$ is equal to $CD$, and $BC$ is common, the two (straight lines) $AB$, $BC$ are equal to the two (straight lines) $DC$, $CB$.¹ And the angle $ABC$ is equal to the angle $BCD$.
Thus, the base $AC$ is equal to the base $BD$, and triangle $ABC$ is equal to triangle $DCB$², and the remaining angles will be equal to the corresponding remaining angles subtended by the equal sides [Prop. 1.4].
Thus, angle $ACB$ is equal to $CBD$.
Also, since the straight line $BC$, (in) falling across the two straight lines $AC$ and $BD$, has made the alternate angles ($ACB$ and $CBD$) equal to one another, $AC$ is thus parallel to $BD$ [Prop. 1.27].
And ($AC$) was also shown (to be) equal to ($BD$).
Thus, straight lines joining equal and parallel (straight lines) on the same sides are themselves also equal and parallel.
(Which is) the very thing it was required to show.

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The Greek text has "$BC$, $CD$", which is obviously a mistake (translator's note). ↩
The Greek text has "$DCB$", which is obviously a mistake (translator's note). ↩