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.

> [!TLDR] > Congruence of triangles satisfies reflexivity (a triangle is congruent to itself), symmetry (if $T_1 \cong T_2$ then $T_2 \cong T_1$), and transitivity (if $T_1 \cong T_2$ and $T_2 \cong T_3$ then $T_1 \cong T_3$). Therefore $R$ 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 $(T_1, T_3) \in R$. > [!WARNING] > A common mistake is to confuse "congruent" with "similar". Congruence requires equal side lengths and equal angles (exact match in size and shape), while similarity only requires equal angles and proportional sides. The relation "is similar to" is also an equivalence relation,

Let T be the set of all triangles in a plane with R a relation in T given by R = \(T₁, T₂): T₁ is congruent to

Worked Examples · Example 2

T

CBSENCERTSubjective· 3mImportance★★★★★

17% · 18/104 Questions

✓ Free question

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

A relation $R$ on a set $A$ is an equivalence relation when it satisfies exactly three properties: it is reflexive, symmetric, and transitive. To prove a given relation is an equivalence relation, you check these three — in this order — one at a time.

Note

Antisymmetry plays no role here; that property belongs to partial orders. For an equivalence relation you need only reflexive, symmetric, transitive.

The three checks

Reflexive — show $(a, a) \in R$ for every $a \in A$ .
Symmetric — assume $(a, b) \in R$ and deduce $(b, a) \in R$ .
Transitive — assume $(a, b) \in R$ and $(b, c) \in R$ , and deduce $(a, c) \in R$ .

If all three hold, $R$ is an equivalence relation. If even one fails, produce a single counterexample and you are done.

A worked template

Let $R$ be defined on $\mathbb{Z}$ by $a\,R\,b \iff a - b$ is divisible by $5$ .

Reflexive: $a - a = 0$ , and $0$ is divisible by $5$ , so $a\,R\,a$ for every integer $a$ . ✓

Symmetric: if $a\,R\,b$ , then $a - b = 5k$ for some integer $k$ . Then $b - a = -5k = 5(-k)$ , also a multiple of $5$ , so $b\,R\,a$ . ✓

Transitive: if $a\,R\,b$ and $b\,R\,c$ , then $a - b = 5k$ and $b - c = 5m$ . Adding, $a - c = 5(k + m)$ , a multiple of $5$ , so $a\,R\,c$ . ✓

All three hold, so $R$ is an equivalence relation.

Tip

Once a relation is proved to be an equivalence relation, it splits $A$ into disjoint equivalence classes — here, the five classes of integers grouped by their remainder on division by $5$ .

Watch out

Reflexivity must hold for every element, not just some. A relation that pairs many elements correctly but misses even one self-pair is not reflexive, and so not an equivalence relation.

Searches like "how to prove a relation is an equivalence relation" and "equivalence relation class 12 maths examples" are common around exam time, reflecting how central this proof technique is to the NCERT/CBSE Class 12 Relations and Functions chapter. Mastering this three-step check (reflexive, symmetric, transitive) also pays off directly in JEE Main and state CET set-theory questions.

←Previous question · 1.2Example 1 — Let A be the set of all students of a boys school. Show that the relat…→Next question · 1.2Example 3 — Let L be the set of all lines in a plane and R be the relation in L de…

Q.Let TTT be the set of all triangles in a plane with RRR a relation in TTT given by R={(T1,T2):T1 is congruent to T2}R = \{(T_1, T_2): T_1 \text{ is congruent to } T_2\}R={(T1​,T2​):T1​ is congruent to T2​}. Show that RRR is an equivalence relation.

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

🎓Unlock everything free for 14 days

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.

Unlock everything free for 14 days