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Math Help - chaleenging one

  1. #1
    Member grgrsanjay's Avatar
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    chaleenging one

    if f,g,h are internal bisectors of angle a of triangle abc,
    Show That \frac{1}{f} \cos \frac{A}{2} + \frac{1}{g} \cos \frac{B}{2}\frac{1}{h}\cos \frac{C}{2} = \frac{1}{a} + \frac{1}{b} +\frac{1}{c}

    where A,B,C are the angles in triangle and a,b,c are sides opposite to the angles A,B,C respectively
    Last edited by grgrsanjay; September 5th 2010 at 07:10 AM.
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  2. #2
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    Let AD, BE and CF be the internal bisectors of angle of triangle ABC.

    Let AD = f. BE = g and CF = h. Now

    \triangle{ABC} = \triangle{ABD} + \triangle{ACD}

    Therefore 1/2*AB*AC*sin(A) = 1/2*AB*AD*sin(A/2) + 1/2*AD*AC*sin(A/2). So

    b*c*sin(A) = c*f*sin(A/2) + f*b*sin(A/2)

    b*c*2sin(A/2)cos(A/2) = c*f*sin(A/2) + f*b*sin(A/2)

    2*b*c*cos(A/2) = f(b+c)

    (1/f)*cos(A/2) = (b+c)/2b*c....(1)

    Similarly

    (1/g)*cos(B/2) = (c+a)/2c*a ...(2) and

    (1/h)*cos(C/2) = (a+b)/2*a*b ....(3)

    Add three equations and simplify.
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  3. #3
    Member grgrsanjay's Avatar
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    yea,thanks got it
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