### In a parallelogram $ABCD$, the bisectors of $\angle A$ and $\angle B$ intersect at $S$, $\angle B$ and $\angle C$ at $R$, $\angle C$ and $\angle D$ at $Q$ and $\angle D$ and $\angle A$ at $P$. What kind of a quadrilateral is $PQRS$?

$Rectangle$

Step by Step Explanation:
1. The situation given in the question is represented by the figure below.
2. We are given that $ABCD$ is a parallelogram. $$\implies DC \parallel AB$$ Also, as the adjacent angles of a parallelogram are supplementary, we have \begin{aligned} & \angle A + \angle D = 180^\circ \\ \implies & \dfrac { 1 } { 2 } \angle A + \dfrac { 1 } { 2 } \angle D = 90^\circ \\ \implies & \angle PAD + \angle ADP = 90^\circ \\ \implies & \angle APD = 90^\circ && \text{[Sum of angles of a triangle is } 180^\circ.] \\ \implies & \angle SPQ = 90^\circ && \text{[Vertically opposite angles.]} \end{aligned} Similarly, $\angle PQR = 90^\circ, \angle QRS = 90^\circ,$ and $\angle PSR = 90^\circ$.

Thus, $PQRS$ is a quadrilateral each of whose angles is $90^\circ$.
Hence, $PQRS$ is a $rectangle$.