Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

(N/A) Given: $A$ parallelogram $ABCD$ in which the bisectors of angles $\angle A, \angle B, \angle C, \angle D$ intersect at points $P, Q, R, S$ to form a quadrilateral $PQRS$.
To Prove: $PQRS$ is a rectangle.
Proof: Since $ABCD$ is a parallelogram,we have $AB \parallel DC$.
Since $AB \parallel DC$ and the transversal $AD$ intersects them at $A$ and $D$ respectively,the sum of consecutive interior angles is $180^{\circ}$.
Therefore,$\angle A + \angle D = 180^{\circ}$.
Dividing by $2$,we get $\frac{1}{2} \angle A + \frac{1}{2} \angle D = 90^{\circ}$.
Since $AS$ and $DS$ are the bisectors of $\angle A$ and $\angle D$ respectively,we have $\angle DAS + \angle ADS = 90^{\circ} \quad \dots(1)$.
In $\triangle DAS$,by the angle sum property of a triangle,$\angle DAS + \angle ADS + \angle ASD = 180^{\circ}$.
Substituting $(1)$ into this,we get $90^{\circ} + \angle ASD = 180^{\circ}$,which implies $\angle ASD = 90^{\circ}$.
Since $\angle PSR$ and $\angle ASD$ are vertically opposite angles,$\angle PSR = \angle ASD = 90^{\circ}$.
Similarly,we can prove that $\angle SRQ = 90^{\circ}$,$\angle RQP = 90^{\circ}$,and $\angle SPQ = 90^{\circ}$.
Since all angles of the quadrilateral $PQRS$ are $90^{\circ}$,$PQRS$ is a rectangle.