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

(N/A) Given: $A$ quadrilateral $ABCD$ in which $AC \perp BD$ and $P, Q, R$ and $S$ are respectively the mid-points of the sides $AB, BC, CD$ and $DA$ of quadrilateral $ABCD$.
To Prove: $PQRS$ is a rectangle.
Proof: In $\Delta ABC$,$P$ and $Q$ are the mid-points of sides $AB$ and $BC$ respectively.
By the Mid-point theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC \dots(1)$
In $\Delta ADC$,$R$ and $S$ are the mid-points of $CD$ and $AD$ respectively.
By the Mid-point theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC \dots(2)$
From $(1)$ and $(2)$,$PQ \parallel SR$ and $PQ = SR$.
Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
Now,let $AC$ and $BD$ intersect at $O$. Let $PQ$ intersect $BD$ at $E$ and $PS$ intersect $AC$ at $G$.
Since $PQ \parallel AC$,$PE \parallel OG$.
Since $PS \parallel BD$,$PG \parallel OE$.
Thus,$PGOE$ is a parallelogram.
Since $AC \perp BD$,$\angle EOG = 90^{\circ}$.
In a parallelogram,opposite angles are equal,so $\angle GPE = \angle EOG = 90^{\circ}$.
Since $PQRS$ is a parallelogram with one angle equal to $90^{\circ}$,$PQRS$ is a rectangle.