Triangle Side-Angle-Side Equality

Theorem
If $2$ triangles have:
 * $2$ sides equal to $2$ sides respectively
 * the angles contained by the equal straight lines equal

they will also have:
 * their third sides equal
 * the remaining two angles equal to their respective remaining angles, namely, those which the equal sides subtend.

Proof


Let $\triangle ABC$ and $\triangle DEF$ be $2$ triangles having sides $AB = DE$ and $AC = DF$, and with $\angle BAC = \angle EDF$.

If $\triangle ABC$ is placed on $\triangle DEF$ such that:
 * the point $A$ is placed on point $D$, and
 * the line $AB$ is placed on line $DE$

then the point $B$ will also coincide with point $E$ because $AB = DE$.

So, with $AB$ coinciding with $DE$, the line $AC$ will coincide with the line $DF$ because $\angle BAC = \angle EDF$.

Hence the point $C$ will also coincide with the point $F$, because $AC = DF$.

But $B$ also coincided with $E$.

Hence the line $BC$ will coincide with line $EF$.

(Otherwise, when $B$ coincides with $E$ and $C$ with $F$, the line $BC$ will not coincide with line $EF$ and two straight lines will enclose a region which is impossible.)

Therefore $BC$ will coincide with $EF$ and be equal to it.

Thus the whole $\triangle ABC$ will coincide with the whole $\triangle DEF$ and thus $\triangle ABC = \triangle DEF$.

The remaining angles on $\triangle ABC$ will coincide with the remaining angles on $\triangle DEF$ and be equal to them.