If two triangles have the two sides equal to two sides respectively, but have the base greater than the base, they will also have the one of the angles contained by the equal straight lines greater than the other.
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,
but it is not;
therefore the angle BAC is not equal to the angle EDF.
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. 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