Question

ABCD is a rhombus and P, Q, R and S are the mid-points of the sides, AB, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Answer 1

In $\Delta$ ABC, P and Q are the mid-points of sides AB and BC respectively.
$\therefore$ PQ || AC and PQ = $\frac{1}{2}$ AC ( Using mid-point theorem) …..(1)
In $\Delta$ADC,
R and S are the mid - points of CD and AD respectively.
$\therefore$ RS || AC and RS = $\frac{1}{2}$ AC ( Using mid-point theorem) …..(2)
From equation (1) and (2) , we obtain
PQ || RS and PQ = RS
Since in quadrilateral PQRs, one pair of opposite sides is equal and parallel to each other , it is a parallelogram.
Let the diagonals of rhombus ABCD intersect each other at point O.
In Quadrilateral OMQN,
MQ || ON ($\because$ PQ || AC)
QN || OM ($\because$ QR || BD)
Therefore, OMQN is a parallelogram.
$\therefore \angle MQN=\angle NOM$
and $\angle PQR=\angle NOM$
However, $\angle NOM={90}^{\circ }$ (Diagonals of a rhombus are perpendicular to each other)
$\therefore \angle PQR={90}^{\circ }$
Clearly , PQRS is a parallelogram having one of its interior angles as ${90}^{\circ }$
Hence, PQRS is a rectangle.