If two triangles have two sides equal to two sides, respectively,  but (one) has the angle encompassed by the equal straight-lines greater than the (corresponding)  angle (in the other), then (the former triangle) will also have a base greater than the base (of the latter).  Let  $ABC$ and $DEF$ be two triangles having the two sides $AB$ and $AC$  equal to the two sides $DE$ and $DF$, respectively. (That is), $AB$ (equal) to $DE$, and  $AC$ to $DF$.  Let them also have the angle at $A$ greater than the angle at $D$.  I say that the base $BC$ is also greater than the base $EF$.  For since angle $BAC$ is greater than angle $EDF$, let (angle) $EDG$, equal to  angle $BAC$,  have been constructed at  the point $D$ on the straight-line $DE$ [Prop.~1.23]. And let $DG$ be made equal to either of $AC$ or $DF$ [Prop.~1.3], and let $EG$ and $FG$ have been joined.  Therefore, since $AB$ is equal to $DE$ and $AC$ to $DG$, the two (straight-lines)  $BA$, $AC$ are equal to the two (straight-lines) $ED$, $DG$, respectively.  Also the angle $BAC$ is equal to the angle $EDG$. Thus, the base $BC$ is equal  to the base $EG$ [Prop.~1.4]. Again, since $DF$ is equal to $DG$, angle $DGF$  is also equal to angle $DFG$ [Prop.~1.5]. Thus, $DFG$ (is) greater than $EGF$.  Thus, $EFG$ is much greater than $EGF$. And since triangle $EFG$ has angle $EFG$  greater than $EGF$, and the greater angle is subtended by the greater side [Prop.~1.19], side $EG$ (is) thus also greater than $EF$. But $EG$ (is) equal to  $BC$. Thus, $BC$ (is) also greater than $EF$.  Thus, if two triangles have two sides equal to two sides, respectively,  but (one) has the angle encompassed by the equal straight-lines greater than the   (corresponding) angle (in the other), then (the former triangle) will also have a base greater than the base (of the latter).  (Which is) the very thing it was required to show.