Show that the relation $R$ defined in the set $A$ of all triangles as $R = \{(T_{1}, T_{2}) : T_{1} \text{ is similar to } T_{2}\}$,is an equivalence relation. Consider three right-angled triangles $T_{1}$ with sides $3, 4, 5$,$T_{2}$ with sides $5, 12, 13$,and $T_{3}$ with sides $6, 8, 10$. Which triangles among $T_{1}, T_{2}$,and $T_{3}$ are related?

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(A) The relation is defined as $R = \{(T_{1}, T_{2}) : T_{1} \text{ is similar to } T_{2}\}$.
$1$. Reflexivity: Every triangle $T_{1}$ is similar to itself. Therefore,$(T_{1}, T_{1}) \in R$. Thus,$R$ is reflexive.
$2$. Symmetry: If $(T_{1}, T_{2}) \in R$,then $T_{1}$ is similar to $T_{2}$. This implies $T_{2}$ is similar to $T_{1}$. Therefore,$(T_{2}, T_{1}) \in R$. Thus,$R$ is symmetric.
$3$. Transitivity: If $(T_{1}, T_{2}) \in R$ and $(T_{2}, T_{3}) \in R$,then $T_{1}$ is similar to $T_{2}$ and $T_{2}$ is similar to $T_{3}$. This implies $T_{1}$ is similar to $T_{3}$. Therefore,$(T_{1}, T_{3}) \in R$. Thus,$R$ is transitive.
Since $R$ is reflexive,symmetric,and transitive,it is an equivalence relation.
Regarding the triangles:
For $T_{1}$ (sides $3, 4, 5$) and $T_{3}$ (sides $6, 8, 10$):
$\frac{3}{6} = \frac{4}{8} = \frac{5}{10} = \frac{1}{2}$.
Since the ratios of corresponding sides are equal,$T_{1}$ is similar to $T_{3}$.
Thus,$T_{1}$ and $T_{3}$ are related.

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