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 Congruence, $\triangle A$ 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 Equality 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
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $3$: Continuity generalized: topological spaces: $3.6$: Homeomorphisms: Examples $3.6.2 \ \text{(e)}$
- in passing
- 1977: Gary Chartrand: Introductory Graph Theory ... (previous) ... (next): Appendix $\text{A}.3$: Equivalence Relations: $(4)$