Let $N$ denote the set of all natural numbers 

and R be the relation on $N$ defined by $\left(\text{a},\text{ b}\right)\text{ R}\left(\text{c},\text{ d}\right)$ if $ad\left(b+c\right)=bc(a+d),$ then $R$ is

We know that for any $a,b\in N$$ab(b+a)=ba(a+b)$

It implies that $(a,b)R(a,b)$

Therefore, the relation $R$ is reflexive. 

Suppose that $\left(a,b\right)R\left(c,d\right)$

it implies that $ad\left(b+c\right)=bc\left(a+d\right)$

we have to show that $\left(c,d\right)R\left(a,b\right)$

This can be written as 

$bc\left(a+d\right)=ad\left(b+c\right)$

it is same as $ad\left(b+c\right)=bc\left(a+d\right)$

Therefore, the relation $R$ is symmetric

Suppose that $\left(a,b\right)R\left(c,d\right)$ and $\left(c,d\right)R\left(e,f\right)$

$\Rightarrow ad(b+c)=bc(a+d)\&cf(d+e)=de(c+f)$

Simplifying the above two equations

we get $af\left(b+e\right)=be\left(a+f\right)$

Hence, $\left(a,b\right)R\left(e,f\right)$

Therefore, the relation $R$ is transitive