If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.
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 angle at A be greater than the angle at D; I say that the base BC is also greater than the base EF.
For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG
equal to the angle BAC; [I. 23] let DG be made equal to either of the two straight lines AC, DF, and let EG, FG be joined. Then, since AB is equal to DE, and AC to DG,
the two sides BA, AC are equal to the two sides ED, DG, respectively;
greater than the angle EGF,
Q. E. D. 1