2. 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.
Given in the question, ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. To Prove, PQRS is a rectangle. Construction, Join AC and BD. Proof: In ΔDRS and ΔBPQ, DS = BQ (Halves of the opposite sides of the rhombus) ∠SDR = ∠QBP (Opposite angles of the rhombus) DR = BP (Halves of the opposite sides of the rhombus) , ΔDRS ≅ ΔBPQ [SAS congruency] RS = PQ [CPCT]———————- (i) In ΔQCR and ΔSAP, RC = PA (Halves of the opposite sides of the rhombus) ∠RCQ = ∠PAS (Opposite angles of the rhombus) CQ = AS (Halves of the opposite sides of the rhombus) , ΔQCR ≅ ΔSAP [SAS congruency] RQ = SP [CPCT]———————- (ii) Now, In ΔCDB, R and Q are the mid points of CD and BC respectively. ⇒ QR || BD also, P and S are the mid points of AD and AB respectively. ⇒ PS || BD ⇒ QR || PS , PQRS is a parallelogram. also, ∠PQR = 90° Now, In PQRS, RS = PQ and RQ = SP from (i) and (ii) ∠Q = 90° , PQRS is a rectangle.
