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

In ΔABC, P and Q are the mid-points of sides AB and BC respectively

AC (Using mid-point theorem) (1)

R and S are the mid-points of CD and AD respectively

RS || AC and RS = 1/2AC (Using mid-point theorem)

∠ PQ || AC and PQ = 1/2AC (2)

In ΔADC,

From equations (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 (PQ || AC)

QN|| OM (QR || BD)

Therefore,

OMQN is a parallelogram

∠MQN = ∠NOM

∠PQR = ∠NOM

However, NOM = 90⁰ (Diagonals of a rhombus are perpendicular to each other)

PQR = 90⁰

Clearly, PQRS is a parallelogram having one of its interior angles as 90⁰

Hence, PQRS is a rectangle