# 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.

In the words of Euclid:

If two triangles have two sides equal to two sides respectively, and have the angles contained by the equal straight lines equal, they will also have the base equal to the base, the triangle will be equal to the triangle, and the remaining angles equal to the remaining angles respectively, 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.

$\blacksquare$

## Historical Note

This proof is Proposition $4$ of Book $\text{I}$ of Euclid's The Elements.