Show that the bisectors of the angles of a parallelogram form a rectangle.

(N/A) Let $P, Q, R$ and $S$ be the points of intersection of the bisectors of $\angle A$ and $\angle B$,$\angle B$ and $\angle C$,$\angle C$ and $\angle D$,and $\angle D$ and $\angle A$ respectively of parallelogram $ABCD$ (see figure).
In $\Delta ASD$,since $DS$ bisects $\angle D$ and $AS$ bisects $\angle A$,we have:
$\angle DAS + \angle ADS = \frac{1}{2} \angle A + \frac{1}{2} \angle D = \frac{1}{2} (\angle A + \angle D)$
Since $\angle A$ and $\angle D$ are interior angles on the same side of the transversal,$\angle A + \angle D = 180^{\circ}$.
Therefore,$\angle DAS + \angle ADS = \frac{1}{2} \times 180^{\circ} = 90^{\circ}$.
Using the angle sum property of a triangle in $\Delta ASD$:
$\angle DAS + \angle ADS + \angle DSA = 180^{\circ}$
$90^{\circ} + \angle DSA = 180^{\circ} \implies \angle DSA = 90^{\circ}$.
Since $\angle PSR$ and $\angle DSA$ are vertically opposite angles,$\angle PSR = 90^{\circ}$.
Similarly,it can be shown that $\angle SPQ = 90^{\circ}$,$\angle PQR = 90^{\circ}$,and $\angle SRQ = 90^{\circ}$.
Since all angles of the quadrilateral $PQRS$ are $90^{\circ}$,$PQRS$ is a rectangle.