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In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that: (iii) ΔAQB ≅ ΔCPD (iv) AQ = CP Q.9(3-4)

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Hello sir i want to know the best solution of the question from exercise 8.1of math of Quadrilaterals chapter of class 9th give me the best and easy for solving this question how i solve it of question no. 9 In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.20). Show that: (iii) ΔAQB ≅ ΔCPD (iv) AQ = CP

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  1. (iii) In ΔAQB and ΔCPD,

    BQ = DP (Given)

    ∠ABQ = ∠CDP (Alternate interior angles)

    AB = CD (Opposite sides of a parallelogram)

    Thus, ΔAQB ≅ ΔCPD [SAS congruency]

    (iv) As ΔAQB ≅ ΔCPD

    AQ = CP [CPCT]

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