$ABCD$ is a rectangle in which diagonal $BD$ bisects $\angle B$. Show that $ABCD$ is a square.

(N/A) Given: $A$ rectangle $ABCD$ in which diagonal $BD$ bisects $\angle B$.
To prove: $ABCD$ is a square.
Proof: $DC \parallel AB$ [Since opposite sides of a rectangle are parallel].
$\Rightarrow \angle 4 = \angle 1$ ... $(1)$ [Alternate interior angles].
Similarly,$\angle 3 = \angle 2$ ... $(2)$ [Alternate interior angles].
And $\angle 1 = \angle 2$ ... $(3)$ [Given,as $BD$ bisects $\angle B$].
From equations $(1)$,$(2)$,and $(3)$,we get $\angle 3 = \angle 4$.
In $\triangle ABD$ and $\triangle CBD$,we have:
$\angle 1 = \angle 2$ [Given]
$BD = BD$ [Common side]
$\angle 3 = \angle 4$ [Proved above]
So,by $ASA$ criterion of congruence,$\triangle ABD \cong \triangle CBD$.
Therefore,$AB = BC$ [Corresponding parts of congruent triangles].
Since $ABCD$ is a rectangle with adjacent sides equal,$ABCD$ is a square.
Hence,proved.