$ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. Show that $ABCD$ is a square.

(N/A) We have a rectangle $ABCD$ such that $AC$ bisects $\angle A$ as well as $\angle C$.
i.e.,$\angle 1 = \angle 4$ and $\angle 2 = \angle 3$ ....... $(1)$
Since a rectangle is a parallelogram,$ABCD$ is a parallelogram.
$\Rightarrow AB \parallel CD$ and $AC$ is a transversal.
$\therefore \angle 2 = \angle 4$ $[\text{Alternate interior angles}]$ .......... $(2)$
From $(1)$ and $(2)$,we have $\angle 3 = \angle 4$.
$\Rightarrow AB = BC$ $[\because \text{sides opposite to equal angles in } \Delta ABC \text{ are equal}]$.
Since $ABCD$ is a rectangle,$AB = CD$ and $BC = AD$.
$\therefore AB = BC = CD = AD$.
Thus,$ABCD$ is a rectangle having all of its sides equal,which means $ABCD$ is a square.