91Ó°ÊÓ

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, $∠A=∠D={90}^{∘}$

Then,

$\begin{array}{c}∠A+∠D={90}^{∘}+{90}^{∘}\\ ={180}^{∘}\end{array}$

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

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

As its angles are bisected,

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

So,

$∠1+∠2={90}^{∘}$

In $\mathrm{Δ}AQD$,

$\begin{array}{c}∠1+∠2+∠Q={180}^{∘}\\ {90}^{∘}+∠Q={180}^{∘}\\ ∠Q={180}^{∘}−{90}^{∘}\\ ={90}^{∘}\end{array}$

Similarly,

$∠P=∠Q=∠R=∠S={90}^{∘}$

Now,

$\begin{array}{c}∠P+∠S={90}^{∘}+{90}^{∘}\\ ={180}^{∘}\end{array}$

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

So,

$PQâˆ\yen RS$.

Also,

$\begin{array}{c}∠P+∠Q={90}^{∘}+{90}^{∘}\\ ={180}^{∘}\end{array}$

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

So,

$PSâˆ\yen RQ$.

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.