Equal triangles which are on equal bases and on the same side are also in the same parallels.

Let $ABC$, $CDE$ be equal triangles on equal bases $BC$, $CE$ and on the same side.I say that they are also in the same parallels.For let $AD$ be joined;I say that $AD$ is parallel to $BE$.For, if not, let $AF$ be drawn through $A$ parallel to $BE$ [Prop. 1.31], and let $FE$ be joined.Therefore the triangle $ABC$ is equal to the triangle $FCE$;for they are on equal bases $BC$, $CE$ and in the same parallels $BE$, $AF$. [Prop. 1.38]But the triangle $ABC$ is equal to the triangle $DCE$;threfore the triangle $DCE$ is also equal to the triangle $FCE$, [C.N. 1]the greater to the less: which is impossible.Therefore $AF$ is not parallel to $BE$.Similarly we can prove that neither is any other straight line except $AD$;therefore $AD$ is parallel to $BE$.Therefore etc.Q.E.D.