Show that the relation R defined in the set A of all triangles as

Solution:

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

R is reflexive since every triangle is similar to itself.

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

$\Rightarrow T_{2}$ is similar to $T_{1}$.

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

∴R is symmetric.

Now,

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

$\Rightarrow T_{1}$ is similar to $T_{2}$ and $T_{2}$ is similar to $T_{3}$.

$\Rightarrow T_{1}$ is similar to $T_{3}$

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

∴ R is transitive.

Thus, R is an equivalence relation.

Now, we can observe that:

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

::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 $T_{3}$.