Question
In the given figure, $\text{ABCD}$ is a quadrilateral. Diagonal $BD$ bisects $\angle B$ and $\angle D$ both. Prove that :
$(i) \triangle \text{ABD} \sim \triangle CBD$
$(ii) \text{AB = BC}$

$(i) \triangle \text{ABD} \sim \triangle CBD$
$(ii) \text{AB = BC}$



