Proof: By Euclid

For if not, being produced, $AB$ and $CD$ will certainly meet together: either in the direction of $B$ and $D$, or (in the direction) of $A$ and $C$ [Def. 1.23] .
Let them have been produced, and let them meet together in the direction of $B$ and $D$ at (point) $G$.
So, for the triangle $GEF$, the external angle $AEF$ is equal to the interior and opposite (angle) $EFG$.
The very thing is impossible [Prop. 1.16].
Thus, being produced, $AB$ and $CD$ will not meet together in the direction of $B$ and $D$.
Similarly, it can be shown that neither (will they meet together) in (the direction of) $A$ and $C$.
But (straight lines) meeting in neither direction are parallel [Def. 1.23] .
Thus, $AB$ and $CD$ are parallel.
Thus, if a straight line falling across two straight lines makes the alternate angles equal to one another then the (two) straight lines will be parallel (to one another).
(Which is) the very thing it was required to show.
∎

Thank you to the contributors under CC BY-SA 4.0!