For let ABC be a triangle having the side AC greater than AB;
I say that the angle ABC is also greater than the angle BCA.

For, since AC is greater than AB, let AD be made equal to AB [I. 3] , and let BD be joined.

Then, since the angle ADB is an exterior angle of the triangle BCD,
it is greater than the interior and opposite angle DCB. [I. 16]

But the angle ADB is equal to the angle ABD,
since the side AB is equal to AD;
therefore the angle ABD is also greater than the angle ACB;
therefore the angle ABC is much greater than the angle ACB.

Therefore etc. Q.E.D.