$ABCD$ is a rectangle 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 rhombus.

(N/A) In a rectangle $ABCD$,$P$ is the mid-point of $AB$,$Q$ is the mid-point of $BC$,$R$ is the mid-point of $CD$,and $S$ is the mid-point of $DA$.
Draw diagonal $AC$.
In $\Delta ABC$,by the Mid-point theorem:
$PQ = \frac{1}{2} AC$ and $PQ \parallel AC$ ......... $(1)$
In $\Delta ACD$,by the Mid-point theorem:
$SR = \frac{1}{2} AC$ and $SR \parallel AC$ ......... $(2)$
From $(1)$ and $(2)$,we get $PQ = SR$ and $PQ \parallel SR$.
Similarly,by joining $BD$,we have $PS = QR$ and $PS \parallel QR$.
Since both pairs of opposite sides of quadrilateral $PQRS$ are equal and parallel,$PQRS$ is a parallelogram.
Now,in $\Delta PAS$ and $\Delta PBQ$:
$\angle A = \angle B = 90^{\circ}$ (Angles of a rectangle)
$AP = BP$ ($P$ is the mid-point of $AB$)
$AS = BQ$ (Halves of equal opposite sides $AD$ and $BC$)
By $SAS$ congruence criterion,$\Delta PAS \cong \Delta PBQ$.
Therefore,$PS = PQ$ (Corresponding parts of congruent triangles).
Since $PQRS$ is a parallelogram with adjacent sides $PS = PQ$,it follows that all sides are equal $(PQ = QR = RS = SP)$.
Thus,$PQRS$ is a rhombus.