Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE and AC to DF; and let the base BC be greater than the base EF;
I say that the angle BAC is also greater than the angle EDF.

For, if not, it is either equal to it or less.

Now the angle BAC is not equal to the angle EDF;
for then the base BC would also have been equal to the base EF, [I. 4]
but it is not;
therefore the angle BAC is not equal to the angle DEF.

Neither again is the angle BAC less than the angle EDF;
for then the base BC would also have been less than the base EF, [I. 24]
but it is not;
therefore the angle BAC is not less than the angle EDF.

But it was proved that it is not equal either;
therefore the angle BAC is greater than the angle EDF.

Therefore, etc. Q.E.D.