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In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(i) ar (BDE) =1/4 ar (ABC) Q.5 (1)

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How i solve the question of class 9th ncert math of Areas of Parallelograms and Triangles chapter of exercise 9.4 of question no 5(1). I think it is very important question of class 9th give me the tricky way for solving this question In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show that:(i) ar (BDE) =1/4 ar (ABC)

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  1. Assume that G and H are the mid-points of the sides AB and AC respectively.

    Join the mid-points with line-segment GH. Here, GH is parallel to third side.

    , BC will be half of the length of BC by mid-point theorem.

    Ncert solutions class 9 chapter 9-33

    ∴ GH =1/2 BC and GH || BD

    ∴ GH = BD = DC and GH || BD (Since, D is the mid-point of BC)

    Similarly,

    GD = HC = HA

    HD = AG = BG

    , ΔABC is divided into 4 equal equilateral triangles ΔBGD, ΔAGH, ΔDHC and ΔGHD

    We can say that,

    ΔBGD = ¼ ΔABC

    Considering, ΔBDG and ΔBDE

    BD = BD (Common base)

    Since both triangles are equilateral triangle, we can say that,

    BG = BE

    DG = DE

    , ΔBDG ΔBDE [By SSS congruency]

    , area (ΔBDG) = area (ΔBDE)

    ar (ΔBDE) = ¼ ar (ΔABC)

    Hence proved

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