MathematicsABCD 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.

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:

    Ncert solutions 9 chapter 8-13

    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: 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)

    Thus, 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)

    Thus, ΔQCR ≅ ΔSAP [SAS congruency]

    RQ = SP [CPCT]———————- (ii)

    Now,

    In ΔCDB

    R and Q are the midpoints of CD and BC, respectively.

    ⇒ QR || BD

    also,

    P and S are the midpoints of AD and AB, respectively.

    ⇒ PS || BD

    ⇒ QR || PS

    Therefore, PQRS is a parallelogram.

    Also, ∠PQR = 90°

     

    In PQRS,

    RS = PQ and RQ = SP from (i) and (ii)

    ∠Q = 90°

    Therefore, PQRS is a rectangle.

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