If two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other.

Let *ABC* and *DEF* be two triangles having two sides *AB* and *AC* equal to two sides *DE* and *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.*

If not, it either equals it or is less.

Now the angle *BAC* does not equal the angle *EDF,* for then the base *BC* would equal the base *EF,* but it is not. Therefore the angle *BAC* does not equal the angle *EDF.*

Neither is the angle *BAC* less than the angle *EDF,* for then the base *BC* would be 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 *if two triangles have two sides equal to two sides respectively, but have the base greater than the base, then they also have the one of the angles contained by the equal straight lines greater than the other.*

Q.E.D.