Question

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

Answer 1

Given $ABCD$ is a rhombus and $PQRS$ are mid point of side $AB,BC,CD$ and $DA$

To Prove: Quadrilateral $PQRS$ is rectangle

Proof: In $△ABC$., $P$ and $Q$ are mid point of $AB$ and $BC$ Since $PQ\parallel AC$ and $PQ=\frac{1}{2}AC$ (mid point theorem) ....(1)

In $△ADC$, $R$ and $S$ are mid point of $DC$ and $AD$ $RS\parallel AC$

and $RS=\frac{1}{2}AC$ (By midpoint theorem)....(2)$

From (1) and (2) we get $(PQ=RS)$

So, in quadrilateral $PQRS$ $PQ\parallel RS$ and $PQ=RS$

Hence $PQRS$ is a parallelogram

Let diagonal of rhombus $ABCD$ intersect at $O$

In quadrilateral $OMQN$

$MQ\parallel ON$ ($\because PQ\parallel AC)$

$QN\parallel OM$ $(\because QR\parallel BD)$

Therefore $OMQN$ is parallelogram

$\angle MQN=\angle NOM$ ie $\angle PQR=\angle NOM$

$\angle NOM={90}^o$ (Diagonal of rhombus are $\perp$ to each other)

$\angle PQR={90}^o$

Hence $PQRS$ is a parallelogram with interior angle ${90}^o$ opposite sides are equal

$\therefore$ $PQRS$ is rectangle