Equal triangles which are on the same base, and on the same side, are also between the same parallels.  Let $ABC$ and $DBC$ be equal triangles which are on the same base $BC$, and on the same side (of it). I say that they are also between the same parallels.  For let $AD$ have been joined. I say that $AD$ and $BC$ are parallel.  For, if not, let $AE$ have been drawn through point A parallel to the straight-line $BC$ [Prop.~1.31], and let $EC$ have been joined. Thus, triangle $ABC$ is equal to triangle $EBC$. For it is on the same base as it, $BC$, and between the same parallels [Prop.~1.37]. But $ABC$ is equal to $DBC$. Thus, $DBC$ is also equal to $EBC$, the greater to the lesser. The very thing is impossible. Thus, $AE$ is not parallel to $BC$. Similarly, we can show that neither (is) any other (straight-line) than $AD$. Thus, $AD$ is parallel to $BC$.  Thus, equal triangles which are on the same base, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show.