Equal triangles which are on equal bases, and on the same side, are also between the same parallels.  Let $ABC$ and $CDE$ be equal triangles on the equal bases $BC$ and $CE$ (respectively), and on the same side (of $BE$). I say that they are also between the same parallels.  For let $AD$ have been joined. I say that $AD$ is parallel to $BE$.  For if not, let $AF$ have been drawn through $A$ parallel to $BE$ [Prop.~1.31], and let $FE$ have been joined. Thus, triangle $ABC$ is equal to triangle $FCE$. For they are on equal bases, $BC$ and $CE$, and between the same parallels, $BE$ and $AF$ [Prop.~1.38]. But, triangle $ABC$ is equal to [triangle] $DCE$. Thus, [triangle] $DCE$ is also equal to triangle $FCE$, the greater to the lesser. The very thing is impossible.  Thus, $AF$ is not parallel to $BE$. Similarly, we can show that neither (is) any other (straight-line) than $AD$. Thus, $AD$ is parallel to $BE$.  Thus, equal triangles which are on equal bases, and on the same side, are also between the same parallels. (Which is) the very thing it was required to show.