If two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in 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 $DF$. And let the base $BC$ be greater than the base $EF$. I say that angle $BAC$ is also greater than $EDF$. <prf>  Thus, $BAC$ is greater than $EDF$.Thus, if two triangles have two sides equal to two sides, respectively, but (one) has a base greater than the base (of the other), then (the former triangle) will also have the angle encompassed by the equal straight-lines greater than the (corresponding) angle (in the latter).