Let $$R$$ be a relation on $${\bf{N}} \times {\bf{N}}$$  defined by $$\left( {a,\,b} \right)R\left( {c,\,d} \right) \Rightarrow ad\left( {b + c} \right) = bc\left( {a + d} \right).\,R$$         is :

Solution :
We observe the following properties :
Reflexivity : Let $$\left( {a,\,b} \right)$$  be an arbitrary element of $${\bf{N}} \times {\bf{N}}$$
$$\eqalign{ & {\text{Then, }}\left( {a,\,b} \right) \in \,{\bf{N}} \times {\bf{N}} \Rightarrow a,\,b\, \in \,{\bf{N}} \cr & \Rightarrow ab\left( {b + a} \right) = ba\left( {a + b} \right) \Rightarrow \left( {a,\,b} \right)R\left( {a,\,b} \right) \cr & {\text{Thus, }}\left( {a,\,b} \right)R\left( {a,\,b} \right){\text{ for all }}\left( {a,\,b} \right) \in \,{\bf{N}} \times {\bf{N}} \cr} $$
So, $$R$$ is reflexive on $${\bf{N}} \times {\bf{N}}$$
Symmetry : Let $$\left( {a,\,b} \right),\,\left( {c,\,d} \right)\, \in \,{\bf{N}} \times {\bf{N}}$$      be such that $$\left( {a,\,b} \right)R\left( {c,\,d} \right).$$
$$\eqalign{ & {\text{Then, }}\left( {a,\,b} \right)R\left( {c,\,d} \right)\, \cr & \Rightarrow ad\left( {b + c} \right) = bc\left( {a + d} \right) \cr & \Rightarrow cb\left( {d + a} \right) = da\left( {c + b} \right) \cr & \Rightarrow \left( {c,\,d} \right)R\left( {a,\,b} \right) \cr & {\text{Thus, }}\left( {a,\,b} \right)R\left( {c,\,d} \right) \cr & \Rightarrow \left( {c,\,d} \right)R\left( {a,\,b} \right){\text{ for all }}\left( {a,\,b} \right)\left( {c,\,d} \right)\, \in \,{\bf{N}} \times {\bf{N}} \cr} $$
So, $$R$$ is symmetric on $${\bf{N}} \times {\bf{N}}$$
Transitivity : Let $$\left( {a,\,b} \right),\,\left( {c,\,d} \right),\,\left( {e,\,f} \right)\, \in \,{\bf{N}} \times {\bf{N}}$$       such that $$\left( {a,\,b} \right)R\left( {c,\,d} \right){\text{ and }}\left( {c,\,d} \right)R\left( {e,\,f} \right)$$
$$\eqalign{ & {\text{Then, }}\left( {a,\,b} \right)R\left( {c,\,d} \right)\, \cr & \Rightarrow ad\left( {b + c} \right) = bc\left( {a + d} \right) \cr & \Rightarrow \frac{{b + c}}{{bc}} = \frac{{a + d}}{{ad}} \cr & \Rightarrow \frac{1}{b} + \frac{1}{c} = \frac{1}{a} + \frac{1}{d}.......({\text{i}}) \cr & {\text{and, }}\left( {c,\,d} \right)R\left( {e,\,f} \right) \Rightarrow cf\left( {d + e} \right) = de\left( {c + f} \right) \cr & \Rightarrow \frac{{d + e}}{{de}} = \frac{{c + f}}{{cf}} \cr & \Rightarrow \frac{1}{d} + \frac{1}{e} = \frac{1}{c} + \frac{1}{f}.......({\text{ii}}) \cr & {\text{Adding (i) and (ii), we get}} \cr & \left( {\frac{1}{b} + \frac{1}{c}} \right) + \left( {\frac{1}{d} + \frac{1}{e}} \right) = \left( {\frac{1}{a} + \frac{1}{d}} \right) + \left( {\frac{1}{c} + \frac{1}{f}} \right) \cr & \Rightarrow \frac{1}{b} + \frac{1}{e} = \frac{1}{a} + \frac{1}{f} \cr & \Rightarrow \frac{{b + e}}{{be}} = \frac{{a + f}}{{af}} \cr & \Rightarrow af\left( {b + e} \right) = be\left( {a + f} \right) \cr & \Rightarrow \left( {a,\,b} \right)R\left( {e,\,f} \right) \cr & {\text{Thus, }}\left( {a,\,b} \right)R\left( {c,\,d} \right){\text{ and }}\left( {c,\,d} \right)R\left( {e,\,f} \right) \cr & \Rightarrow \left( {a,\,b} \right)R\left( {e,\,f} \right){\text{ for all }}\left( {a,\,b} \right),\,\left( {c,\,d} \right),\,\left( {e,\,f} \right)\, \in \,{\bf{N}} \times {\bf{N}} \cr} $$
So, $$R$$ is transitive on $${\bf{N}} \times {\bf{N}}.$$
Hence, $$R$$ being reflexive, symmetric and transitive, is an equivalence relation on $${\bf{N}} \times {\bf{N}}.$$