Let $A$ be the set of all triangles in a plane. Show that the relation $\mathrm{R}=\left\{\left(\Delta_{1}, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ is an equivalence relation on $\mathrm{A}$.

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Let $A$ be the set of all triangles in a plane. Show that the relation $\mathrm{R}=\left\{\left(\Delta_{1}, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ is an equivalence relation on $\mathrm{A}$ .

Let $A$ be the set of all triangles in a plane. Show that the relation $\mathrm{R}=\left\{\left(\Delta_{1}, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ is an equivalence relation on $\mathrm{A}$ .

Solution:

Suppose $\mathrm{R}=\left\{\left(\Delta 1, \Delta_{2}\right): \Delta_{1} \sim \Delta_{2}\right\}$ be a relation defined on A. (As given)
If $\mathrm{R}$ is Reflexive, Symmetric and Transitive, therefore $\mathrm{R}$ is an equivalence relation.
So now,

Reflexivity:
Suppose $\Delta$ be an arbitrary element of $A$
we have,
$\Delta \sim \Delta$ since, every triangle is similar to itself.
$\Rightarrow(\Delta, \Delta) \in \mathrm{R}$ and $\Delta \in \mathrm{A}$
Therefore, $R$ is reflexive.

Symmetric:
Suppose $\Delta_{1}$ and $\Delta_{2} \in A$ , such that $\left(\Delta 1, \Delta_{2}\right) \in R$
$\begin{array}{l} \Rightarrow \Delta 1 \sim \Delta 2 \\ \Rightarrow \Delta 2 \sim \Delta_{1} \\ \Rightarrow\left(\Delta 2, \Delta_{1}\right) \in \mathrm{R} \end{array}$
Therefore, $\mathrm{R}$ is symmetric

Transitivity:
Suppose $\Delta_{1}, \Delta_{2}, \Delta_{3} \in \mathrm{A}$ such that $\left(\Delta_{1}, \Delta_{2}\right) \in \mathrm{R}$ and $\left(\Delta_{2}, \Delta_{3}\right) \in \mathrm{R}$ $\Rightarrow \Delta_{1} \sim \Delta_{2}$ and $\Delta_{2} \sim \Delta_{3}$
$\Rightarrow \Delta_{1} \sim \Delta_{3}$
$\Rightarrow\left(\Delta_{1}, \Delta_{3}\right) \in \mathrm{R}$
Therefore, $\mathrm{R}$ is transitive.
As a result, $\mathrm{R}$ is an equivalence relation.