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XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ΔABE) = ar(ΔACF) Q.8

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Yesterday i was doing the question from class 9th ncert book of math of Areas of Parallelograms and Triangles chapter of exercise 9.3  What is the easiest way for solving it because i was not able to do this question please help me for solving this question XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and F respectively, show that ar(ΔABE) = ar(ΔACF)

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  1. Ncert solutions class 9 chapter 9-17

    Given,

    XY || BC, BE || AC and CF || AB

    To show,

    ar(ΔABE) = ar(ΔAC)

    Proof:

    BCYE is a || gm as ΔABE and ||gm BCYE are on the same base BE and between the same parallel lines BE and AC.

    ∴,ar(ABE) =  ½ ar(BCYE) … (1)

    Now,

    CF || AB and XY || BC

    ⇒ CF || AB and XF || BC

    ⇒ BCFX is a || gm

    As ΔACF and || gm BCFX are on the same base CF and in-between the same parallel AB and FC .

    ∴,ar (ΔACF)= ½  ar (BCFX) … (2)

    But,

    ||gm BCFX and || gm BCYE are on the same base BC and between the same parallels BC and EF.

    ∴,ar (BCFX) = ar(BCYE) … (3)

    From (1) , (2) and (3) , we get

    ar (ΔABE) = ar(ΔACF)

    ⇒ ar(BEYC) = ar(BXFC)

    As the parallelograms are on the same base BC and in-between the same parallels EF and BC–(iii)

    Also,

    △AEB and ||gm BEYC are on the same base BE and in-between the same parallels BE and AC.

    ⇒ ar(△AEB) = ½ ar(BEYC) — (iv)

    Similarly,

    △ACF and || gm BXFC on the same base CF and between the same parallels CF and AB.

    ⇒ ar(△ ACF) = ½ ar(BXFC) — (v)

    From (iii), (iv) and (v),

    ar(△ABE) = ar(△ACF)

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