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Updated on : 2022-09-05

Solution

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$ABCD$ is a rhombus.

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

Join $AC$ and $BD.$

In $△DBC,$

$R$ and $Q$ are the mid-points of $DC$ and $CB$

$∴$ $RQ∥DB$ [ By mid-point theorem ]

$∴$ $MQ∥ON$ ---- ( 1 ) [ Parts of $RQ$ and $DB$ ]

Now, in $△ACB,$

$P$ and $Q$ are the mid-points of $AB$ and $BC$

$∴$ $AC∥PQ$ [ By midpoint theorem ]

$∴$ $OM∥NQ$ ---- ( 2 ) [ $OM$ and $NQ$ are the parts of $AC$ and $PQ$ ]

From equation ( 1 ) and ( 2 )

$⇒$ $MQ∥ON$

$⇒$ $ON∥NQ$

Since, each pair of opposite side is parallel.

$∴$ $ONQM$ is a parallelogram.

In $ONQM,$

$⇒$ $∠MON=90_{o}$ [ Diagonals of rhombus bisect each other ]

$⇒$ $∠MON=∠PQR$ [ Opposite angles of parallelogram are equal ]

$∴$ $∠PQR=90_{o}.$

Hence, $PQ⊥QR$ ---- Hence proved.

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