Question

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.

Answer 1

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

∠ PQ || AC and PQ = $\frac{1}{2}$

AC (Using mid-point theorem) ... (1)

In ΔADC,

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

∠RS || AC and RS = $\frac{1}{2}$ AC (Using mid-point theorem) ... (2)

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 = 90° (Diagonals of a rhombus are perpendicular to each other) ∠PQR = 90°

Clearly, PQRS is a parallelogram having one of its interior angles as 90º.

Hence, PQRS is a rectangle.