Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ is congruent to $\left. \mathrm{T} _{2}\right\}$ . Show that $\mathrm{R}$ is an equivalence relation.
Let $\mathrm{T}$ be the set of all triangles in a plane with $\mathrm{R}$ a relation in $\mathrm{T}$ given by $\mathrm{R} =\left\{\left( \mathrm{T} _{1}, \mathrm{T} _{2}\right): \mathrm{T} _{1}\right.$ is congruent to $\left. \mathrm{T} _{2}\right\}$ . Show that $\mathrm{R}$ is an equivalence relation.
$\mathrm{R}$ is reflexive, since every triangle is congruent to it self.
Further, $\left( \mathrm{T} _{1}, \,\mathrm{T}_{2}\right) \in \mathrm{R} \Rightarrow \mathrm{T} _{1}$ is congruent to $\mathrm{T} _{2} \Rightarrow \mathrm{T} _{2}$ is congruent to $\mathrm{T} _{1} \Rightarrow\left( \mathrm{T} _{2}, \mathrm{T} _{1}\right) \in \mathrm{R} .$
Hence, $\mathrm{R}$ is symmetric.
Moreover, $\left( \mathrm{T} _{1},\, \mathrm{T} _{2}\right),\left( \mathrm{T} _{2}, \,\mathrm{T} _{3}\right) \in \mathrm{R} \Rightarrow \mathrm{T} _{1}$ is congruent to $\mathrm{T} _{2}$ and $\mathrm{T} _{2}$ is congruent to $\mathrm{T} _{3} \Rightarrow \mathrm{T} _{1}$ is congruent to $\mathrm{T} _{3} \Rightarrow\left( \mathrm{T} _{1}, \,\mathrm{T} _{3}\right) \in \mathrm{R}$.
Therefore, $\mathrm{R}$ is an equivalence relation.
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