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3CD is a Rhombus and P, Q, R and S arepoints of sides AB, BC, CD, DA respectithat FTPORS i< a Rectanole 5 |
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Answer» Given- 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,AC and BD are joined. 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)Thus, ΔDRS ≅ ΔBPQ by SAS congruence condition.RS = PQ by 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)Thus, ΔQCR ≅ ΔSAP by SAS congruence condition.RQ = SP by CPCT --- (ii)Now,In ΔCDB,R and Q are the mid points of CD and BC respectively.⇒ QR || BDalso,P and S are the mid points of AD and AB respectively.⇒ PS || BD⇒ QR || PSThus, PQRS is a parallelogram.also, ∠PQR = 90°Now,In PQRS,RS = PQ and RQ = SP from (i) and (ii)∠Q = 90°Thus, PQRS is a rectangle. |
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