# Congruence of Triangles is Equivalence Relation

## Theorem

Let $S$ denote the set of all triangles in the plane.

Let $\triangle A \cong \triangle B$ denote the relation that $\triangle A$ is congruent to $\triangle B$.

Then $\cong$ is an equivalence relation on $S$.

## Proof

Checking in turn each of the criteria for equivalence:

### Reflexivity

Let $\triangle A$ be a triangle.

By definition, by Triangle Side-Side-Side Equality, $\triangle A$ is trivially congruent to itself.

Thus $\cong$ is seen to be reflexive.

$\Box$

### Symmetry

Let $\triangle A \cong \triangle B$.

Then:

- all the angles contained by the sides of $\triangle A$ are equal to the angles contained by the sides of $\triangle B$.

It follows directly that:

- all the angles contained by the sides of $\triangle B$ are equal to the angles contained by the sides of $\triangle A$.

That is:

- $\triangle B \cong \triangle A$

Thus $\cong$ is seen to be symmetric.

$\Box$

### Transitivity

Let:

- $\triangle A \cong \triangle B$
- $\triangle B \cong \triangle C$

Then:

- all the angles contained by the sides of $\triangle A$ are equal to the angles contained by the sides of $\triangle B$.

and:

- all the angles contained by the sides of $\triangle B$ are equal to the angles contained by the sides of $\triangle C$.

From Equals is Equivalence Relation, it follows that:

- all the angles contained by the sides of $\triangle A$ are equal to the angles contained by the sides of $\triangle C$.

That is:

- $\triangle A \cong \triangle C$

Thus $\cong$ is seen to be transitive.

$\Box$

$\cong$ has been shown to be reflexive, symmetric and transitive.

Hence by definition it is an equivalence relation.

$\blacksquare$

## Sources

- 1977: Gary Chartrand:
*Introductory Graph Theory*... (previous) ... (next): Appendix $\text{A}.3$: Equivalence Relations: $(4)$