Triangle Side-Angle-Side Equality

Theorem
If two triangles have:
 * two sides equal to two 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 two 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 space 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.

Note
This is Proposition 4 of Book I of Euclid's "The Elements".