Question

$ABCD$ 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.

Answer 1

Step $1:$ Drawing the diagram:

$ABCD$ is a rectangle,

$P,Q,R$ and $S$ are mid-points of the sides $AB,BC,CD$ and $DA$ respectively.

Join diagonals $AC$ and $BD$ which intersect at $O$.

$PQ$ and $OB$ intersect at $M$ and

$RQ$ and $OC$ intersect at $N$ .

Step $2:$ Proving $PQRS$ is a parallelogram:

In, $∆ADC$,

$S$ is the midpoint of $AD$ and $R$ is the midpoint of $CD$.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, $SR\parallel AC$ and $SR=\frac{1}{2}AC$…………………..(i)

In, $∆ABC$,

$P$ is the midpoint of $AB$ and $Q$ is the midpoint of $BC$.

Therefore, by mid point theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half ot it.

Therefore, $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$……………..(ii)

From (i) and (ii), we get

Therefore, $PQ=SR$

We have, $SR\parallel AC$ and $PQ\parallel AC$

$\therefore SR\parallel PQ$ (Lines parallel to same line are parallel to each other)

And, also $PQ=SR$.

$\therefore PQRS$ is a parallelogram because a pair of opposite side of quadrilateral $PQRS$ is equal and parallel.

Step $3:$Proving $PQRS$ is a rectangle:

As, $MQ\parallel ON$ and $OM\parallel NQ$ (Parts of parallel lines as $PQRS$ is a parallelogram)

$\angle MON=\angle MQN$ (Opposite angles of parallelogram are equal)

$\Rightarrow \angle MON=\angle MQN=\angle PQR$

But, $\angle MON=90°$ (Diagonals of rhombus bisect each other at right angles)

$\therefore \angle PQR=90°$

Now, in parallelogram $PQRS$,

$PQ=SR$ and $PS=RQ$ (Opposite sides of parallelogram are equal)

and $\angle PQR=90°$

Therefore, $PQRS$ is a rectangle.