According to the question ,

Given set S = {…All triangles in plane….}

And R = {(∆₁ , ∆₂) : ∆₁ , ∆₂∈ S and ∆₁ ≡ ∆₂}

Formula

For a relation R in set A

Reflexive

The relation is reflexive if (a , a) ∈ R for every a ∈ A

Symmetric

The relation is Symmetric if (a , b) ∈ R , then (b , a) ∈ R

Transitive

Relation is Transitive if (a , b) ∈ R & (b , c) ∈ R , then (a , c) ∈ R

Equivalence

If the relation is reflexive , symmetric and transitive , it is an equivalence relation.

Check for reflexive

Consider , (∆₁, ∆₁)

∴ We know every triangle is congruent to itself.

(∆₁, ∆₁) ∈ R all ∆₁ ∈ S

Therefore , R is reflexive ……. (1)

Check for symmetric

(∆₁ , ∆₂) ∈ R then ∆₁ is congruent to ∆₂

(∆₂ , ∆₁) ∈ R then ∆₂ is congruent to ∆₁

Both the equation are the same and therefore will always be true.

Therefore , R is symmetric ……. (2)

Check for transitive

Let ∆₁, ∆₂, ∆₃ ∈ S such that (∆₁, ∆₂) ∈ R and (∆₂, ∆₃) ∈ R

Then (∆₁, ∆₂)∈R and (∆₂, ∆₃)∈R

⇒∆₁ is congruent to ∆_2, and ∆₂ is congruent to ∆₃

⇒∆₁ is congruent to ∆₃

∴(∆₁, ∆₃) ∈ R

Therefore , R is transitive ……. (3)

Now , according to the equations (1) , (2) , (3)

Correct option will be (D)