Draw a rectangle and bisect its angles. The bisectors intersect to form what special kind of quadrilateral.

Here, ABCD is a rectangle and its bisector intersect to form a quadrilateral PQRS.

ABCD is a rectangle and each angle of a rectangle is $90°$.

So, $\angle A=\angle D={90}^{\circ }$

Then,

$\begin{array}{c}\angle A+\angle D={90}^{\circ }+{90}^{\circ }\\ ={180}^{\circ }\end{array}$

Multiply both sides by $\frac{1}{2}$,

$\begin{array}{c}\frac{1}{2}\left(\angle A+\angle D\right)=\frac{1}{2}\left({180}^{\circ }\right)\\ \frac{1}{2}\angle A+\frac{1}{2}\angle D={90}^{\circ }\end{array}$

As its angles are bisected,

$\angle 1=\frac{1}{2}\angle A$and $\angle 2=\frac{1}{2}\angle D$.

So,

$\angle 1+\angle 2={90}^{\circ }$

In $\Delta AQD$,

$\begin{array}{c}\angle 1+\angle 2+\angle Q={180}^{\circ }\\ {90}^{\circ }+\angle Q={180}^{\circ }\\ \angle Q={180}^{\circ }-{90}^{\circ }\\ ={90}^{\circ }\end{array}$

Similarly,

$\angle P=\angle Q=\angle R=\angle S={90}^{\circ }$

Now,

$\begin{array}{c}\angle P+\angle S={90}^{\circ }+{90}^{\circ }\\ ={180}^{\circ }\end{array}$

If a transversal intersects two lines such that a pair of interior angles is supplementary, then the lines are parallel.

So,

$PQ\parallel RS$.

Also,

$\begin{array}{c}\angle P+\angle Q={90}^{\circ }+{90}^{\circ }\\ ={180}^{\circ }\end{array}$

If a transversal intersects two lines such that a pair of interior angles are supplementary, then the lines are parallel.

So,

$PS\parallel RQ$.

Therefore, PQRS is a parallelogram as both sides are parallel.

Also, each of the angles in the parallelogram is equal to $90°$.

So,

PQRS is a rectangle.

Therefore, a quadrilateral is a rectangle.