$P, Q, R$ and $S$ are respectively the mid-points of sides $AB, BC, CD$ and $DA$ of quadrilateral $ABCD$ in which $AC = BD$ and $AC \perp BD$. Prove that $PQRS$ is a square.

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AC = BD$ and $AC \perp BD$. $P, Q, R$ and $S$ are respectively the mid-points of sides $AB, BC, CD$ and $DA$.
To prove: $PQRS$ is a square.
Proof:
$1$. In $\triangle ABC$,$P$ and $Q$ are mid-points of $AB$ and $BC$ respectively. By the Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$.
$2$. Similarly,in $\triangle ADC$,$S$ and $R$ are mid-points of $AD$ and $CD$ respectively. Thus,$SR \parallel AC$ and $SR = \frac{1}{2} AC$.
$3$. Since $PQ = SR$ and $PQ \parallel SR$,$PQRS$ is a parallelogram.
$4$. In $\triangle ABD$,$P$ and $S$ are mid-points of $AB$ and $AD$ respectively. Thus,$PS = \frac{1}{2} BD$.
$5$. Since $AC = BD$ (given),we have $PQ = \frac{1}{2} AC = \frac{1}{2} BD = PS$. Thus,$PQ = PS$.
$6$. Since $PQRS$ is a parallelogram with adjacent sides equal $(PQ = PS)$,it is a rhombus.
$7$. Since $AC \perp BD$ and $PQ \parallel AC$ and $PS \parallel BD$,it follows that $PQ \perp PS$. Thus,$\angle SPQ = 90^{\circ}$.
$8$. $A$ rhombus with one angle equal to $90^{\circ}$ is a square. Therefore,$PQRS$ is a square.