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in triangle abc angle b = 2angle c.d is a point ob bc such that ad bisects angle bac and ab = cd.prove that angle bac = 72

  • maths - ,

    make a sketch.
    let angle C = x , then angle B = 2x
    also, let angle BAD = y = angle CAD , (Angle A was bisected by AD)
    then in triangle ABC , 2x + 2y + x = 180
    y = (180-3x)/2 = 90 - 3x/2

    by exterior angle theorem,
    angle BDA = x+y and angle CDA = 2x+y
    let AB = m = DC , (given)

    by the sine law:
    m/sin(x+y) = AD/sin 2x and m/siny = AD/sinx
    m = ADsin(x+y)/sin 2x and m = ADsiny/sinx

    so ADsin(x+y)/sin2x = ADsiny/sinx
    divide by AD
    sin(x+y)/sin2x = siny/sinx
    sin(x + 90-3x/2)/(2sinxcosx) = siny/sinx
    sin(90 - x/2)/(2cosx) = siny , after dividing by sinx
    cross-multiply
    sin(90-x/2) = 2sinxcosx
    sin(90-x/2) = sin 2x
    so 90 - x/2 = 2x
    times 2
    180 - x = 4x
    180=5x
    x=36
    then y = 90-3(36)/2 = 36
    and angle A = 2y = 72°

  • maths - ,

    There is a wonderful field of math lying ahead from this problem.
    notice that cos 36° = .809016.. which is 1/2 of the golden ratio of (1+√5)/2 or 1.61803....

    Look up the "pentagon"
    Draw the central triangles, your triangle is one of these.

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