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

$\text { (i) } \operatorname{In} \triangle ABD \ \triangle CBD$
$ \angle 3=\angle 4$
$\angle 1=\angle 2$
$\therefore \triangle ABD \sim \triangle CBD$
$\text { (ii) } \triangle ABD \cong \triangle CBD$
$\therefore AB = BC $
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