Given that,

R be a relation on T defined as aRb if a is congruent to b ∀ a, b ∈ T

Now,

aRa ⇒ a is congruent to a, which is true since every triangle is congruent to itself.

⇒ (a,a) ∈ R ∀ a ∈ T

⇒ R is reflexive.

Let aRb ⇒ a is congruent to b

⇒ b is congruent to a

⇒ bRa

∴ (a,b) ∈ R ⇒ (b,a) ∈ R ∀ a, b ∈ T

⇒ R is symmetric.

Let aRb ⇒ a is congruent to b and bRc ⇒ b is congruent to c

⇒ a is congruent to c

⇒ aRc

∴ (a,b) ∈ R and (b,c) ∈ R ⇒ (a,c) ∈ R ∀ a, b,c ∈ T

⇒ R is transitive.

Hence, R is an equivalence relation.