Proof: (Euclid)

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.
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