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 $EDF$.  For if not, ($BAC$) is certainly either equal to, or less than, ($EDF$). In fact, $BAC$ is not equal to $EDF$. For then the base $BC$ would also have been equal to the base $EF$ [Prop.~1.4]. But it is not. Thus, angle $BAC$ is not equal to $EDF$. Neither, indeed, is $BAC$ less than $EDF$. For then the base $BC$ would also have been less than the base $EF$ [Prop.~1.24]. But it is not. Thus, angle $BAC$ is not less than $EDF$. But it was  shown that ($BAC$ is) not equal (to $EDF$) either. 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). (Which is) the very thing it was required to show.