ABCD is a rhombus and P,Q,R,S …

In Δ ABC, P and Q are the mid-points of sides AB and BC respectively. ∴ PQ || AC and PQ = 1 2 AC ( Using mid-point theorem) …..(1) In Δ ADC, R and S are the mid - points of CD and AD respectively. ∴ RS || AC and RS = 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 ( ∵ PQ || AC) QN || OM ( ∵ QR || BD) Therefore, OMQN is a parallelogram. ∴ ∠ M Q N = ∠ N O M and ∠ P Q R = ∠ N O M However, ∠ N O M = 90 ∘ (Diagonals of a rhombus are perpendicular to each other) ∴ ∠ P Q R = 90 ∘ Clearly , PQRS is a parallelogram having one of its interior angles as 90 ∘ Hence, PQRS is a rectangle.