Q.Let $T$ be the set of all triangles in a plane with $R$ a relation in $T$ given by $R = \{(T_1, T_2): T_1 \text{ is congruent to } T_2\}$. Show that $R$ is an equivalence relation.

Concept: Relation Properties — an equivalence relation must be reflexive, symmetric, and transitive.

Step 1 — Reflexive: Every triangle $T_1$ is congruent to itself (by the identity congruence). So $(T_1, T_1) \in R$ for all $T_1 \in T$.

Step 2 — Symmetric: If $T_1$ is congruent to $T_2$, then by definition $T_2$ is congruent to $T_1$ (congruence is mutual). Hence $(T_1, T_2) \in R \implies (T_2, T_1) \in R$.

Step 3 — Transitive: If $T_1 \cong T_2$ and $T_2 \cong T_3$, then $T_1 \cong T_3$ (congruence is transitive). So $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$ imply $(T_1, T_3) \in R$.

Since all three properties hold, $R$ is an equivalence relation.

The question asks us to show that the relation "is congruent to" on the set of all triangles is an equivalence relation. An equivalence relation must satisfy three properties: reflexivity, symmetry, and transitivity. Each of these corresponds to a basic fact about geometric congruence — facts you already know from your study of triangles.

Let’s check them one by one.

1. Reflexivity: A relation $R$ is reflexive if every element is related to itself.

   For any triangle $T_1$, we have $T_1 \cong T_1$ because every triangle is congruent to itself (by the identity mapping — same side lengths, same angles).

   Hence $(T_1, T_1) \in R$ for all $T_1 \in T$.

2. Symmetry: $R$ is symmetric if whenever $(T_1, T_2) \in R$, then $(T_2, T_1) \in R$.

   If $T_1$ is congruent to $T_2$, then by definition there exists an isometry (a combination of translation, rotation, reflection) mapping $T_1$ onto $T_2$. The inverse of that isometry maps $T_2$ onto $T_1$, so $T_2 \cong T_1$.

   Therefore $(T_2, T_1) \in R$ whenever $(T_1, T_2) \in R$.

3. Transitivity: $R$ is transitive if whenever $(T_1, T_2) \in R$ and $(T_2, T_3) \in R$, then $(T_1, T_3) \in R$.

   If $T_1 \cong T_2$ and $T_2 \cong T_3$, then there exist isometries $f$ and $g$ such that $f(T_1) = T_2$ and $g(T_2) = T_3$. The composition $g \circ f$ is also an isometry, and $(g \circ f)(T_1) = T_3$. Hence $T_1 \cong T_3$.

   So $(T_1, T_3) \in R$.

Since all three properties hold, $R$ is an equivalence relation.

Method: Proving a relation is an equivalence relation

Use this standard three-part proof whenever you must show a relation $R$ on a set is an equivalence relation.

Steps

Step 1: Prove reflexivity

Show $(a,a)\in R$ for every $a$ in the set, by verifying the defining property holds for an element compared with itself.

Step 2: Prove symmetry

Assume $(a,b)\in R$ and deduce $(b,a)\in R$, using that the underlying relation runs both ways (here, "is congruent to" is a mutual property).

Step 3: Prove transitivity

Assume $(a,b)\in R$ and $(b,c)\in R$, and deduce $(a,c)\in R$ by chaining the two facts (for congruence, compose the two matchings).

Conclude: all three hold, so $R$ is an equivalence relation. The same skeleton works for any equivalence-relation proof — only the justification of each step changes with the definition of $R$.

Common Mistakes

Mistake 1: Confusing congruence with similarity

Why it's wrong: similarity only needs equal angles and proportional sides; congruence needs equal sides and angles. Using a similarity argument (scaling) to justify congruence is a different relation. Correct approach: justify each property using congruence (identical size and shape) — reflexive by identity, symmetric because congruence is mutual, transitive by chaining.

Mistake 2: Declaring "it's obviously an equivalence relation" without checking all three properties

Why it's wrong: a board answer must verify reflexivity, symmetry and transitivity separately; asserting the conclusion earns no marks and hides the reasoning. Correct approach: write one short justification per property before concluding.