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geometry

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A pentagon ABCDE. EG drawn parallel to DA meets BA produced at G and CF drawn parallel to DB meets AB produced at F. Prove that the area of pentagon ABCDE is equal to the area of triangle GDF.

  • geometry - ,

    Lets concentrate on the quadrilateral ADEG.
    let the intersection of AE and DB be K
    GE || AD (given)
    So triangles ADE and ADG have the same base AD and the same height , since parallel lines are equal distances apart.
    So triangle ADE = triangle ADG
    but ADK is common to both, so subtracting ...
    triangle AKG = triangle DEK

    similarly by letting FD and BC intersect at L we can show that triangle FLB = triangle CDL

    pentagon ABCDE
    = triangle BLD + triangle LCD + triangle ABD + triangle DEK + triangle ADK
    = triangle BLD + triangle FLB + triangle ABD + triangle GAK + triangle ADk
    = triangle FDG

  • geometry - ,

    tri FDG

  • geometry - ,

    In given figure X and Y are the mid point of AC and AB respectively QP parallel BC and CYQ and BXP are straight line. Prove that ar Triangle ABP =ar triangle ACQ

  • geometry - ,

    Since Y and X are the nid points of AB and AC ; by mid point theorem XY || BC. Now consider the ∆ BCQ and ∆ CBP their areas are equal cz they are on same base BC and between the same parallels
    : AR(BCQ) = AR(BCP)
    Let the intersection point of QC and BP be O
    : add AR(AQOP) both sides.....
    Therefore AR (ABP) =AR(ACQ)

  • geometry - ,

    Jabbba jabba kappale mayathu

  • geometry - ,

    This question is wrong

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