Given $R= \{ (T_1,T_2):T_1$ is similar to $T_2 \}$:

For any triangle $T_1, (T_1,T_1)\in R$ as every triangle is similar to itself. Hence $R$ is reflexive.

If any triangle $T_1$ is similar to $T_2$ then $T_2$ will be similar to $T_1$. Hence it follows that $(T_1,T_2) \in R \Rightarrow (T_2,T_1) \in R$. Therefore, $R$ is symmetric.

If $T_1$ is similar to $T_2$ and $T_2$ is similar to $T_3$ then $T_1$ is similar to $T_3$ also .

Therefore, $(T_1,T_2), (T_2,T_3) \in R \Rightarrow (T_1,T_3) \in R$. Hence, $R$ is transitive.

Since $R$ is reflexive, symmetric and transitive, it is an equivalance relation

Given three right angle triangles $T_1$ with sides $3, 4, 5, T_2$ with sides $5, 12, 13$, and $T_3$, with sides $6, 8, 10$.

We can see that the circles of triangle $T_1$ and $T_3$ are proportional because: $ \frac{3}{6}=\frac{4}{8}=\frac{5}{10}=(\frac{1}{2})$

Therefore $T_1$ and $T_3$ are similar and related.

However, the circles of any two other triangles for e.g. $T_1$ and $T_2$ are not propotional as evidenced by: $ \frac{3}{5} \neq \frac{4}{12} \neq \frac{5}{13}$