Let R be the relation "is congruent to" on the set of all triangles...

Let $S$ denote the set of all triangles in a plane. Let $R$ be the relation on $S$ defined by \left(\Delta_1, \Delta_2\right) \inR\Rightarrow$triangle$\Delta_1 \cong \Delta_2$: (i) Let any triangle$\Delta \in S$, we have$\Delta_1 \cong \Delta_2 \Rightarrow(\Delta, \Delta) \in R \forall \Delta \in S \Rightarrow R$is reflexive on (ii) Let$\Delta_1, \Delta_2 \in \mathrm{~S}$, such that \left(\Delta_1, \Delta_2\right) \in R$, then $\Delta_1 \cong \Delta_2 \Rightarrow \Delta_2 \cong \Delta_1 \Rightarrow\left(\Delta_2, \Delta_1\right) \in R \Rightarrow R$ is symmętric (iii) Again, let $\Delta_1, \Delta_2, \Delta_3 \in \mathrm{~S}$ such that \left(\Delta_1, \Delta_2\right)\in R$and \left(\Delta_2, \Delta_3\right) \in R \therefore \Delta_1 \cong \Delta_2 \cong \Delta_3\therefore\left(\Delta_1, \Delta_2\right) \in R\Rightarrow R$ is transitive. $\therefore \mathrm{R}$ is an equivalence relation.