Triangle Side-Side-Side Equality

Theorem

Let two triangles have all three sides equal.

Then they also have all three angles equal.

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

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 line $BC$ on $EF$.

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

Now 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, in contradiction to 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 is Proposition 8 of Book I of Euclid's The Elements.