Triangle Side-Side-Side Congruence/Proof 1

Theorem

Let two triangles have all $3$ sides equal.

Then they also have all $3$ angles equal.

Thus two triangles whose sides are all equal are themselves congruent.

In the words of Euclid:

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

(The Elements: Book $\text{I}$: Proposition $8$)

Proof

Let $\triangle ABC$ and $\triangle DEF$ be two triangles such that:

$AB = DE$

$AC = DF$

$BC = EF$

Suppose $\triangle ABC$ were superimposed over $\triangle DEF$ so that point $B$ is placed on point $E$ and the side $BC$ on $EF$.

Then $C$ will coincide with $F$, as $BC = EF$ and so $BC$ coincides with $EF$.

Aiming for a contradiction, suppose $BA$ does not coincide with $ED$ and $AC$ does not coincide with $DF$.

Then they will fall as, for example, $EG$ and $GF$.

Thus there will be two pairs of straight line segments constructed on the same line segment, on the same side as it, meeting at different points.

But from Proposition $7$: Two Lines Meet at Unique Point, these points are coincident.

It follows by Proof by Contradiction that:

$BA$ coincides with $ED$

$AC$ coincides with $DF$.

Therefore $\angle BAC$ coincides with $\angle EDF$ and therefore:

$\angle BAC = \angle EDF$

The same argument can be applied mutatis mutandis to the other two sides.

Thus we have shown that all corresponding angles are equal.

$\blacksquare$

Historical Note

This proof is Proposition $8$ of Book $\text{I}$ of Euclid's The Elements.

Sources

• 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 1 (2nd ed.) ... (previous) ... (next): Book $\text{I}$. Propositions