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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.

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Solution

## In ΔABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = AC (Using mid-point theorem) ... (1) In ΔADC, R and S are the mid-points of CD and AD respectively. ∴ RS || AC and RS = 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.

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