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

Let $ABC$, $DBC$ be equal triangles which are on the same base $BC$ and on the same side of it;[I say that they are also in the same parallels.]And [For] let $AD$ be joined;I say that $AD$ is parallel to $BC$.For, if not, let $AE$ be drawn through the point $A$ parallel to the straight line $BC$, [Prop. 1.31]and let $EC$ be joined.Therefore the triangle $ABC$ is equal to the triangle $EBC$;for it is on the same base $BC$ with it and in the same parallels. [Prop. 1.37]But $ABC$ is equal to $DBC$;therefore $DBC$ is also equal to $EBC$, [C.N. 1]the greater to the less: which is impossible.Therefore $AE$ is not parallel to $BC$.Similarly we can prove that neither is any other straight line except $AD$;therefore $AD$ is parallel to $BC$.Therefore etc.Q.E.D.