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

For, if not, AC is either equal to AB or less.

Now AC is not equal to AB;
for then the angle ABC would also have been equal to the angle ACB; [I. 5]
but it is not;
therefore AC is not equal to AB.

Neither is AC less than AB,
for then the angle ABC would also have been less than the angle ACB; [I. 18]
but it is not;
therefore AC is not less than AB.

And it was proved that it is not equal either.
Therefore AC is greater than AB.

Therefore etc. Q.E.D.