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In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar(△ACB) = ar(△ACF) Q.11(1)

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How i find the best tricky way for solving the question of class 9th of Areas of Parallelograms and Triangles chapter of class 9th give me the best and easy method for this question no.11 of exercise 9.3 In Fig. 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that (i) ar(△ACB) = ar(△ACF)

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    1. △ACB and △ACF lie on the same base AC and between the same parallels AC and BF.

    ∴ar(△ACB) = ar(△ ACF)

    1. ar(△ACB) = ar(△ACF)

    ⇒ ar(△ACB)+ar(△ACDE) = ar(△ACF)+ar(△ACDE)

    ⇒  ar(ABCDE) = ar(△AEDF)

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