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 $EF$.  <prf>  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).