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

Let *ABC* and *DEF* be two triangles having the two sides *AB* and *AC* equal to the two sides *DE* and *DF* respectively, so that *AB* equals *DE,* and *AC* equals *DF,* and let the angle at *A* be greater than the angle at *D.*

I say that the base *BC* is greater than the base *EF.*

Since the angle *BAC* is greater than the angle *EDF,* construct the angle *EDG* equal to the angle *BAC* at the point *D* on the straight line *DE.* Make *DG* equal to either of the two straight lines *AC* or *DF.* Join *EG* and *FG.*

Since *AB* equals *DE,* and *AC* equals *DG,* the two sides *BA* and *AC* equal the two sides *ED* and *DG,* respectively, and the angle *BAC* equals the angle *EDG,* therefore the base *BC* equals the base *EG.*

Again, since *DF* equals *DG,* therefore the angle *DGF* equals the angle *DFG.* Therefore the angle *DFG* is greater than the angle *EGF.*

Therefore the angle *EFG* is much greater than the angle *EGF.*

Since *EFG* is a triangle having the angle *EFG* greater than the angle *EGF,* and side opposite the greater angle is greater, therefore the side *EG* is also greater than *EF.*

But *EG* equals *BC,* therefore *BC* is also greater than *EF.*

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

Q.E.D.