If two triangles have the two sides equal to two sides respectively, but one has the angles contained by the equal straight lines greater than the other, it 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 $BAC$; [Prop. 1.23]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;and the angle $BAC$ is equal to the angle $EDG$;therefore the base $BC$ is equal to the base $EG$. [Prop. 1.4]Again, since $DF$ is equal to $DG$,the angle $DGF$ is is also equal to the angle $DFG$; [Prop. 1.5]therefore the angle $DFG$ is greater than the angle $EGF$.Therefore the angle $EFG$ is much greater than the angle $EGF$.And, since $EFG$ is a triangle having the angle $EFG$ greater than the angle $EGF$,and the greater angle is subtended by the greater side, [Prop. 1.19]the side $EG$ is also greater than $EF$.But $EG$ is equal to $BC$.Therefore $BC$ is also greater than $EF$.Therefore etc.Q.E.D.