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.

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Solution

## 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 getTherefore, $PQ=SR$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)$⇒\angle MON=\angle MQN=\angle PQR$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.

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