Question

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

Answer 1

Let us draw the figure with given condition as

In $△ABC$, $P$ and $Q$ are the mid- points of $AB$ and $BC$.

$PQ\parallel AC$ and by using mid - point theorem

$PQ=\frac{1}{2}AC$ ........(1)

Similarly, in $△ADC$, $R$ and $S$ are the mid- points of $CD$ and $AD$.

$SR\parallel AC$ and by using mid point theorem

$SR=\frac{1}{2}AC$ .........(2)

From (1) and (2), we get
$PQ\parallel RS$ and $PQ=SR$

Now, in quadrilateral $PQRS$ its one pair of opposite sides $PQ$ and $SR$ is equal and parallel.

Therefore,

$PQRS$ is a parallelogram

$AB=BC$ (Sides of a rhombus)

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}BC$

$PB=BQ$

$\angle 3=\angle 4$

Now, in $△APS$ and $△CQR$, we have

$AP=CQ$ (Halves of equal sides $AB,BC$)

$AS=CR$ (Halves of equal sides $AD,CD$)

$PS=QR$ (Opp. sides of parallelogram $PQRS$)

Therefore, $△APS\cong △CQR$ $\mid$ using SSS Congruency Theorem

$\angle 1=\angle 2$ (Corresponding parts of congruent triangles are equal)

Now , $\angle 1+\angle SPQ+\angle 3={180}^o$ (Linear pair axiom)

Therefore, $\angle 1+\angle SPQ+\angle 3=\angle 2+\angle PQR+\angle 4$

But, $\angle 1=\angle 2$ and $\angle 3=\angle 4$

Therefore,

$\angle SPQ=\angle PQR$ .....(3)

Since, $SP\parallel RQ$ and $PQ$ intersects them

Therefore, $\angle SPQ+\angle PQR={180}^o$....(4) (Since consecutive interior angles are supplementary)

From (3) and (4), we get

$\angle PQR+\angle PQR={180}^o$

$2\angle PQR={180}^o$

$\angle PQR={90}^o$

$\angle SPQ=\angle PQR={90}^o$

Thus, $PQRS$ is a parallelogram whose one angle $\angle SPQ={90}^o$.

Hence $PQRS$ is a rectangle.