Triangle Side-Angle-Side Equality
Theorem
If two triangles have:
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.
$\blacksquare$
Historical Note
This is Proposition 4 of Book I of Euclid's The Elements.
