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, I.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.

References

Footnotes