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.

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

Then:
 * all the sides of $\triangle A$ are equal to the sides of $\triangle B$


 * 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 sides of $\triangle B$ are equal to the sides of $\triangle A$


 * 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.

Transitivity
Let:
 * $\triangle A \cong \triangle B$
 * $\triangle B \cong \triangle C$

Then:
 * all the sides of $\triangle A$ are equal to the sides of $\triangle B$


 * 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 sides of $\triangle B$ are equal to the sides of $\triangle C$


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

From Equality is Equivalence Relation, it follows that:


 * all the sides of $\triangle A$ are equal to the sides of $\triangle C$


 * 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.

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

Hence by definition it is an equivalence relation.