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Prove that the parallelogram circumscribing a circle is a rhombus. Q.11

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How i prove the parallelogram of question no.11 of ncert class 10th book of exercise 10.2, please give me the simple way for to prove the parallelogram  Prove that the parallelogram circumscribing a circle is a rhombus.

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  1. Consider a parallelogram ABCD which is circumscribing a circle with a center O. Now, since ABCD is a parallelogram, AB = CD and BC = AD.

    Ncert solutions class 10 chapter 10-13

    From the above figure, it is seen that,

    (i) DR = DS

    (ii) BP = BQ

    (iii) CR = CQ

    (iv) AP = AS

    These are the tangents to the circle at D, B, C, and A respectively.

    Adding all these we get,

    DR+BP+CR+AP = DS+BQ+CQ+AS

    By rearranging them we get,

    (BP+AP)+(DR+CR) = (CQ+BQ)+(DS+AS)

    Again by rearranging them we get,

    AB+CD = BC+AD

    Now, since AB = CD and BC = AD, the above equation becomes

    2AB = 2BC

    ∴ AB = BC

    Since AB = BC = CD = DA, it can be said that ABCD is a rhombus.

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