$R =\{\left( T _{1}, T _{2}\right): T _{1}$  is similar to $T _{2}\}$

$R$ is reflexive since every triangle is similar to itself.

Further,

If $\left(T_{1},\, T_{2}\right) \in R,$ then $T_{1}$ is similar to $T_{2} .$

$\Rightarrow T _{2}$ is similar to $T _{1}$

$\Rightarrow\left(T_{2}, T_{1}\right) \in R$

$\therefore R$ is symmetric.

Now,

Let $\left(T_{1}, T_{2}\right),\left(T_{2}, T_{3}\right) \in R$

$\Rightarrow$ Ti is similar to $T _{2}$ and $T _{2}$ is similar to $T _{3}$.

$\Rightarrow T _{1}$ is similar to $T_3$

$\Rightarrow\left(T_{1},\, T_{3}\right) \in R$

$\therefore R$ is transitive.

Thus, $R$ is an equivalence relation.

Now,

We can observe that $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}\left(=\frac{1}{2}\right)$

$\therefore$ The corresponding sides of triangles $T _{1}$ and $T _{3}$ are in the same ratio.

Then, triangle $T _{1}$ is similar to triangle $T _{3}$.

Hence, $T _{1}$ is related to $T _{3}$.