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.

Step 1: Reflexivity

For a relation to be reflexive, every element must be related to itself. In this case, we need to show that any triangle $T_1$ is congruent to itself. This is true by definition, as a triangle is always congruent to itself.

Step 2: Symmetry

For a relation to be symmetric, if $T_1$ is related to $T_2$, then $T_2$ must be related to $T_1$. If triangle $T_1$ is congruent to triangle $T_2$, it logically follows that triangle $T_2$ is also congruent to triangle $T_1$. Congruence is a symmetric property.

Step 3: Transitivity

For a relation to be transitive, if $T_1$ is related to $T_2$ and $T_2$ is related to $T_3$, then $T_1$ must be related to $T_3$. If triangle $T_1$ is congruent to triangle $T_2$, and triangle $T_2$ is congruent to triangle $T_3$, then triangle $T_1$ must be congruent to triangle $T_3$. This is a fundamental property of congruence.

Step 4: Conclusion

Since the relation R satisfies all three properties: reflexivity, symmetry, and transitivity, it is an equivalence relation.

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