In the set of triangles in a plane the relation 'is similar to' is an equivalence relation. Prove .

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Solution

Let T be the set of all triangles in a plane.
We are given that $T_{1} R T_{2}$
$\Rightarrow T_{1} : T_{2}$ for all $T_{1}, T_{2} \in T$
Reflexive : Let $T_{1} \in T$ such that $T_{1} R T_{1}$ .
Then $T_{1} R T \Rightarrow T_{1} : T_{1}$ every triangle is similar to itself.
So, : is reflexive on T.
Symmetric: Let $T_{1}, T_{2} \in T$ such that $T_{1} R T_{2}$
Then $T_{1} R T_{2} \Rightarrow T_{1} : T_{2} \Rightarrow T_{2} : T_{1}$
So, r is symmetric on T.
Transitive: Let $T_{1}, T_{2}, T_{3} \in T$ such that $T_{1} R T_{2}, T_{2} R T_{3}$ .
Then, $T_{1} R T_{2} \Rightarrow T_{1} : T_{2}$ and $T_{2} : T_{3}$ implies that $T_{1} : T_{3}$
So, $R$ is transitive on $T$ .
Hence, $R$ is an equivalence relation on $T$ .

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