In quadrilateral $ABCD$,$AC = BD$ and $AC \perp BD$. If $P, Q, R$ and $S$ are the midpoints of the sides $AB, BC, CD$ and $DA$ respectively,prove that $PQRS$ is a square.

(N/A) $1$. In $\triangle ABC$,$P$ and $Q$ are midpoints of $AB$ and $BC$. By the Midpoint Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
$2$. In $\triangle ADC$,$S$ and $R$ are midpoints of $AD$ and $CD$. By the Midpoint Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
$3$. Thus,$PQ \parallel SR$ and $PQ = SR$. Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
$4$. Similarly,in $\triangle ABD$,$PS \parallel BD$ and $PS = \frac{1}{2} BD$. In $\triangle BCD$,$QR \parallel BD$ and $QR = \frac{1}{2} BD$.
$5$. Since $AC = BD$ (given),then $\frac{1}{2} AC = \frac{1}{2} BD$,which implies $PQ = SR = PS = QR$. Thus,$PQRS$ is a rhombus.
$6$. Since $AC \perp BD$,and $PQ \parallel AC$ and $PS \parallel BD$,it follows that $PQ \perp PS$. Therefore,$\angle QPS = 90^{\circ}$.
$7$. $A$ rhombus with one angle equal to $90^{\circ}$ is a square. Hence,$PQRS$ is a square.