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, [I. 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. [I. 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.