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**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.

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