Given: A quadrilateral ABCD whose diagonals AC and BD are perpendicular to each other at O. P,Q,R and S are mid points of side AB, BC, CD and DA respectively are joined are formed quadrilateral PQRS. To Prove : PQRS is a rectangle. Proof : In △ABC, P and Q are mid - points of AB and BC respectively. ∴ PQ || AC and PQ = 1 / 2 AC ---- (i) [mid point theorem] Further, in △ACD, R and S are mid points of CD and DA respectively. ∴ SR || AC and SR = 1 / 2 AC --- (ii) [mid point theorem] From (i) and (ii) , we have PQ || SR and PQ = SR Thus , one pair of opposite sides of quadrilateral PQRS are parallel and equal . ∴ PQRS is a parallelogram . Since PQ || AC ⇒ PM || NO In △ABD, P and S are mid points of AB and AD respectively . ∴ PS || BD [mid point theorem] ⇒ PN || MO ∴ Opposite sides of quadrilateral PMON parallel . ∴ PMON is a parallelogram . ∴ ∠MPN = ∠MON [opposite angles of || gm are equal] But ∠MON = 90° [give] ∴ ∠MPN = 90° ⇒ ∠QPS = 90° Thus, PQRS is a parallelogram whose one angle is 90°. ∴ PQRS is a rectangle.