Proof: By Euclid
(related to Proposition: 1.04: "SideAngleSide" Theorem for the Congruence of Triangle)
 For if triangle $ABC$ is applied to triangle $DEF$,^{1} the point $A$ being placed on the point $D$, and the straight line $AB$ on $DE$, then the point $B$ will also coincide with $E$, on account of $AB$ being equal to $DE$.
 So (because of) $AB$ coinciding with $DE$, the straight line $AC$ will also coincide with $DF$, on account of the angle $BAC$ being equal to $EDF$.
 So the point $C$ will also coincide with the point $F$, again on account of $AC$ being equal to $DF$.
 But, point $B$ certainly also coincided with point $E$, so that the base $BC$ will coincide with the base $EF$.
 For if $B$ coincides with $E$, and $C$ with $F$, and the base $BC$ does not coincide with $EF$, then two straight lines will encompass an area.
 The very thing^{2} is impossible [Post. 1] .
 Thus, the base $BC$ will coincide with $EF$, and will be equal to it [C.N. 4] .
 So the whole triangle $ABC$ will coincide with the whole triangle $DEF$, and will be equal to it [C.N. 4] .
 And the remaining angles will coincide with the remaining angles, and will be equal to them [C.N. 4] .
 (That is) $ABC$ to $DEF$, and $ACB$ to $DFE$ [C.N. 4] .
 Thus, if two triangles have two sides equal to two sides, respectively, and have the angle(s) enclosed by the equal straight line equal, then they will also have the base equal to the base, and the triangle will be equal to the triangle, and the remaining angles subtended by the equal sides will be equal to the corresponding remaining angles.
 (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"
Footnotes