$ABCD$ is a rhombus and $P, Q, R$ and $S$ are mid-points of the sides $AB, BC, CD$ and $DA$ respectively. Show that the quadrilateral $PQRS$ is a rectangle.

(N/A) Given: $ABCD$ is a rhombus. $P, Q, R, S$ are mid-points of sides $AB, BC, CD, DA$ respectively.
To prove: $PQRS$ is a rectangle.
Construction: Join $AC$ and $BD$.
Proof:
$1$. In $\Delta ABC$,$P$ and $Q$ are mid-points of $AB$ and $BC$. By Mid-point Theorem,$PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ ... $(1)$.
$2$. In $\Delta ADC$,$S$ and $R$ are mid-points of $AD$ and $CD$. By Mid-point Theorem,$SR \parallel AC$ and $SR = \frac{1}{2} AC$ ... $(2)$.
$3$. From $(1)$ and $(2)$,$PQ \parallel SR$ and $PQ = SR$. Since one pair of opposite sides is equal and parallel,$PQRS$ is a parallelogram.
$4$. In a rhombus,diagonals bisect each other at $90^{\circ}$. Let $AC$ and $BD$ intersect at $O$. Thus,$AC \perp BD$.
$5$. Since $PQ \parallel AC$ and $QR \parallel BD$,the angle between $PQ$ and $QR$ must be $90^{\circ}$ because the angle between $AC$ and $BD$ is $90^{\circ}$.
$6$. Since $PQRS$ is a parallelogram with one angle equal to $90^{\circ}$,$PQRS$ is a rectangle.