Let A be the set of all triangles in a plane. Show that the relation

R = {(∆1, ∆2) : ∆1 ~ ∆2} is an equivalence relation on A.


Let R = {(∆1, ∆2) : ∆1 ~ ∆2} be a relation defined on A.


Now,


R is Reflexive if (Δ, Δ) R Δ A


We observe that for each Δ A we have,


Δ ~ Δ since, every triangle is similar to itself.


(Δ, Δ) R Δ A


R is reflexive.


R is Symmetric if (1, ∆2) R (2, ∆1) R 1, ∆2 A


Let (1, ∆2) R 1, ∆2 A


1 ~ ∆2


2 ~ ∆1


(2, ∆1) R


R is symmetric


R is Transitive if (1, ∆2) R and (2, ∆3) R (1, ∆3) R 1, ∆2, 3 A


Let (1, ∆2) R and ((2, ∆3) R 1, ∆2, 3 A


1 ~ ∆2 and 2 ~ ∆3


1 ~ ∆3


(1, ∆3) R


R is transitive.


Since R is reflexive, symmetric and transitive, it is an equivalence relation on A.


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