Proof: By Euclid

For if triangle $ABC$ is applied to triangle $DEF$, the point $B$ being placed on point $E$, and the straight line $BC$ on $EF$, then point $C$ will also coincide with $F$, on account of $BC$ being equal to $EF$.
So (because of) $BC$ coinciding with $EF$, (the sides) $BA$ and $CA$ will also coincide with $ED$ and $DF$ (respectively).
For if base $BC$ coincides with base $EF$, but the sides $AB$ and $AC$ do not coincide with $ED$ and $DF$ (respectively), but miss like $EG$ and $GF$ (in the above figure), then we will have constructed upon the same straight line, two other straight lines equal, respectively, to two (given) straight lines, and (meeting) at a different point on the same side (of the straight line), but having the same ends.
But (such straight lines) cannot be constructed [Prop. 1.7].
Thus, the base $BC$ being applied to the base $EF$, the sides $BA$ and $AC$ cannot coincide with $ED$ and $DF$ (respectively).
Thus, they will coincide.
So the angle $BAC$ will also coincide with angle $EDF$, and will be equal to it [C.N. 4] .
Thus, if two triangles have two sides equal to two side, respectively, and have the base equal to the base, then they will also have equal the angles encompassed by the equal straight lines.
(Which is) the very thing it was required to show.
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