The straight lines joining equal and parallel straight lines (at the extremities which are) in the same directions (respectively) are themselves also equal and parallel.

Let $AB$, $CD$ be equal and parallel, and let the straight lines $AC$, $BD$ join them (at the extremities which are) in the same directions (respectively);I say that $AC$, $BD$ are also equal and parallel.Let $BC$ be joined.Then, since $AB$ is parallel to $CD$, and $BC$ has fallen upon them,the alternate angles $ABC$, $BCD$ are equal to one another. [Prop. 1.29]And, since $AB$ is equal to $CD$, and $BC$ is common,the two sides $AB$, $BC$ are equal to the two sides $DC$, $CB$;and the angle $ABC$ is equal to the angle $BCD$;therefore the base $AC$ is equal to the base $BD$,and the triangle $ABC$ is equal to the triangle $DCB$,and the remaining angles will be equal to the remaining angles respectively, namely those which the equal sides subtend; [Prop. 1.4]therefore the angle $ACB$ is equal to the angle $CBD$.And, since the straight line $BC$ falling on the two straight lines $AC$, $BD$ has made the alternate angles equal to one another,$AC$ is parallel to $BD$. [Prop. 1.27]And it was also proved equal to it.Therefore etc.Q.E.D.