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.  For let the straight-line $EF$, falling across the two straight-lines $AB$ and $CD$, make the alternate angles $AEF$ and $EFD$ equal to one another. I say that $AB$ and $CD$ are parallel.  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.