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 [I. 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.

But the triangle ABC is equal to the triangle DCE;
therefore 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.