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

Hint: For showing $PQRS$ as a rhombus, we must show that all its sides are equal and opposite sides are parallel. So for proving $PQRS$ as rhombus, we must prove that $PQ\parallel RS,PS\parallel RQ$ and also $PQ = PS = RS = RQ$. We can use the midpoint theorem for the triangle.

Complete step-by-step answer:
As we know that $ABCD$ is a rectangle so we can say that $AB = CD,BC = AD$ and also $AB\parallel CD,AC\parallel AD$ and also each and every angle is $90^\circ $
So $\angle A = \angle B = \angle C = \angle D = 90^\circ $

Now let us assume that $P,Q,R,S$ are the midpoints of the sides $AB,BC,CD,DA$
Now we can prove that $PQRS$ is a rhombus and we know that the parallelogram with equal sides is called a rhombus.
So constructing $AC$ we get two different triangles $\Delta ABC,\Delta ADC$
In$\Delta ABC$, $P,Q$ are the midpoints of the sides $AB,BC$ respectively.
So we can write by the midpoint theorem that
$PQ\parallel AC{\text{ and }}PQ = \dfrac{1}{2}AC$$ - - - - - - (1)$
In$\Delta ADC$, $R,S$ are the midpoints of the sides $CD,AD$ respectively.
So we can write by the midpoint theorem that
$RS\parallel AC{\text{ and RS}} = \dfrac{1}{2}AC$$ - - - - - - (2)$
Therefore in (1), we get that \[PQ\parallel AC\]
And in the equation (2) we get that \[RS\parallel AC\]
Therefore all are parallel. So \[PQ\parallel AC\parallel RS\]
Also we get that $PQ = \dfrac{1}{2}AC,RS = \dfrac{1}{2}AC$
So we get that $PQ = RS$
Therefore we get that $PQ = RS$ and also $PQ\parallel RS$. Hence if opposite sides are parallel and equal then it is a parallelogram
Now in $\Delta APS,\Delta BPQ$
We know that $P$ is the midpoint of $AB$
So $AP = BP$
And as $ABCD$ is a rectangle $\angle A = \angle B = 90^\circ $
And we know that opposite sides are equal so $AD = BC$ and we also can write that $\dfrac{{AD}}{2} = \dfrac{{BC}}{2}$
So we can write that $AS = BQ$ and $S,Q$ are midpoints of $AD,BC$
Therefore we get that $AP = BP$
$\angle PAS = \angle PBQ$
$AS = BQ$ as proved above
So we can write that
$\Delta APS \cong \Delta BPQ$ (by SAS congruency)
Since both the triangles are congruent, then corresponding parts are also congruent. So we can write
$PS = PQ$
And we proved that $PQRS$ is a parallelogram. So $PQ = RS$ and $PS = RQ$
So we proved $PS = PQ$
So we can write that this means that all the sides are equal.
Therefore $PQRS$ is a rhombus because a parallelogram with all the sides equal is a rhombus.

So $PQRS$ is a rhombus.

Note: Every square, rectangle, rhombus is a parallelogram but every parallelogram is not a rectangle, square and a rhombus. Rhombus and square both have all sides equal but in square each angle is $90^\circ $ but not in the case of rhombus.