ABCD is a rhombus with P,Q,R as mid-points of AB, BC and CD. Prove that $PQ\perp QR$.

$ABCD$ 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 $CB$

$\therefore$  $RQ\parallel DB$                [ By mid-point theorem ]

$\therefore$  $MQ\parallel ON$       ---- ( 1 )   [ Parts of $RQ$ and $DB$ ]

Now, in $△ACB,$

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

$\therefore$  $AC\parallel PQ$              [ By midpoint theorem ]

$\therefore$  $OM\parallel NQ$       ---- ( 2 )     [ $OM$ and $NQ$ are the parts of $AC$ and $PQ$ ]

From equation ( 1 ) and ( 2 )

$\Rightarrow$  $MQ\parallel ON$

$\Rightarrow$  $ON\parallel NQ$

Since, each pair of opposite side is parallel.

$\therefore$  $ONQM$ is a parallelogram.

In $ONQM,$

$\Rightarrow$  $\angle MON=90^o$            [ Diagonals of rhombus bisect each other ]

$\Rightarrow$  $\angle MON=\angle PQR$       [ Opposite angles of parallelogram are equal ]

$\therefore$  $\angle PQR=90^o.$

Hence, $PQ\perp QR$             ---- Hence proved.