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  1. Asked: In:

    Diagonals AC and BD of a quadrilateral ABCD intersect at O in such a way that ar (AOD) = ar (BOC). Prove that ABCD is a trapezium.

    Best Answer
    kamal shahu

    ar(AOD) = ar(BOC) [∵ Given] adding ar(AOB) both of sides ar(AOD) + ar(AOB) = ar(BOC) + ar(AOB) ⇒ ar(ABD) = ar(ABC) ΔABC and ΔABC are on the same base AB and ar(ABD) = ar(ABC). Therefore, AB ∥ DC [∵ Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.] HeRead more

    ar(AOD) = ar(BOC) [∵ Given]
    adding ar(AOB) both of sides
    ar(AOD) + ar(AOB) = ar(BOC) + ar(AOB)
    ⇒ ar(ABD) = ar(ABC)
    ΔABC and ΔABC are on the same base AB and ar(ABD) = ar(ABC).
    Therefore, AB ∥ DC
    [∵ Triangles on the same base (or equal bases) and having equal areas lie between

    the same parallels.]
    Hence, ABCD is a trapezium.

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  2. Asked: In:

    D and E are points on sides AB and AC respectively of Δ ABC such that ar (DBC) = ar (EBC). Prove that DE || BC.

    Best Answer
    kamal shahu

    ΔDBC and ΔEBC are on the same base BC and ar (DBC) = ar(EBC). Therefore, DE ∥ BC [∵ Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.]

    ΔDBC and ΔEBC are on the same base BC and ar (DBC) = ar(EBC).
    Therefore, DE ∥ BC
    [∵ Triangles on the same base (or equal bases) and having equal areas lie between the same parallels.]

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  3. Asked: In:

    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 (ABE) = ar (ACF)

    Best Answer
    kamal shahu

    In quadrilateral BCYE, BE ∥ CY [∵ BE ∥ AC] BC ∥ EY [∵ BC ∥ XY] Therefore, BCYE is a parallelogram. Triangle ABE and Parallelogram BCYE are on the same base BE and between same parallels, BE ∥ AC. Hence, ar(ABE) = (1/2)ar(BCYE) ...(1) [∵ If a parallelogram and a triangle are on the same base and betwRead more

    In quadrilateral BCYE, BE ∥ CY [∵ BE ∥ AC]
    BC ∥ EY [∵ BC ∥ XY]
    Therefore, BCYE is a parallelogram.
    Triangle ABE and Parallelogram BCYE are on the same base BE and between
    same parallels, BE ∥ AC.
    Hence, ar(ABE) = (1/2)ar(BCYE) …(1)
    [∵ If a parallelogram and a triangle are on the same base and between the same parallels, than area of the triangle is half the area of the parallelogram.]
    Similarly, triangle ACF and parallelogram BCFT are on the same base CF and Between same parallels CF ∥ AB.
    Hence, ar(ACF) = (1/2)ar(BCFX) …(2)
    And, ar(BCYE) = ar(BCFX) …(3)
    [∵ On the same base (BC) and between same parallels (BC ∥ EF), area of parallograms are equal]
    From the equation (1), (2) and (3), ar (ABE) = ar(ACF)

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