Triangle Side-Side-Side Equality

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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

Euclid-I-8.png

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.

This contradicts the theorem Two Lines Meet at Unique Point.

Therefore $BA$ coincides with $ED$ and $AC$ coincides with $DF$.

Therefore $\angle BAC$ coincides with $\angle EDF$ and is equal to it.


The same argument can be applied to the other two sides, and thus we show that all corresponding angles are equal.

$\blacksquare$


Historical Note

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


Sources