Straight-lines joining equal and parallel (straight-lines) on the same sides are  themselves also equal and parallel.  Let $AB$ and $CD$ be equal and parallel (straight-lines), and let the straight-lines $AC$ and $BD$ join them on the same sides. I say that $AC$ and $BD$ are also equal and parallel.  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) $DC$, $CB$.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 $DCB$, 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.