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Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint : From A and C, draw perpendiculars to BD.] Q.6

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Please guide me the best way for solving the question of class 9th math of Areas of Parallelograms and Triangles chapter of exercise 9.4 of math of question no. 6 What is the best way for solving this question, please guide me Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (APB)×ar (CPD) = ar (APD)×ar (BPC). [Hint : From A and C, draw perpendiculars to BD.]

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  1. Given:

    The diagonal AC and BD of the quadrilateral ABCD, intersect each other at point E.

    Construction:

    From A, draw AM perpendicular to BD

    From C, draw CN perpendicular to BD

    Ncert solutions class 9 chapter 9-35

    To Prove,

    ar(ΔAED) ar(ΔBEC) = ar (ΔABE) ×ar (ΔCDE)

    Proof,

    ar(ΔABE) = ½ ×BE×AM………….. (i)

    ar(ΔAED) = ½ ×DE×AM………….. (ii)

    Dividing eq. ii by i , we get,

    Ncert solutions class 9 chapter 9-36

    ar(AED)/ar(ABE) = DE/BE…….. (iii)

    Similarly,

    ar(CDE)/ar(BEC) = DE/BE ……. (iv)

    From eq. (iii) and (iv) , we get

    ar(AED)/ar(ABE) = ar(CDE)/ar(BEC)

    , ar(ΔAED)×ar(ΔBEC) = ar(ΔABE)×ar (ΔCDE)

    Hence proved.

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