Proof: (Euclid)

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.
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