PROPOSITION 40.¶

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 [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. [I. 38]

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.]