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.
$R$ is reflexive,since every triangle is congruent to itself.
Further,$(T_1, T_2) \in R \implies T_1 \text{ is congruent to } T_2 \implies T_2 \text{ is congruent to } T_1 \implies (T_2, T_1) \in R$.
Hence,$R$ is symmetric.
Moreover,$(T_1, T_2) \in R$ and $(T_2, T_3) \in R \implies T_1 \text{ is congruent to } T_2$ and $T_2 \text{ is congruent to } T_3 \implies T_1 \text{ is congruent to } T_3 \implies (T_1, T_3) \in R$.
Therefore,$R$ is an equivalence relation.
Further,$(T_1, T_2) \in R \implies T_1 \text{ is congruent to } T_2 \implies T_2 \text{ is congruent to } T_1 \implies (T_2, T_1) \in R$.
Hence,$R$ is symmetric.
Moreover,$(T_1, T_2) \in R$ and $(T_2, T_3) \in R \implies T_1 \text{ is congruent to } T_2$ and $T_2 \text{ is congruent to } T_3 \implies T_1 \text{ is congruent to } T_3 \implies (T_1, T_3) \in R$.
Therefore,$R$ is an equivalence relation.
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