if f,g,h are internal bisectors of angle a of triangle abc,

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where A,B,C are the angles in triangle and a,b,c are sides opposite to the angles A,B,C respectively

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- September 3rd 2010, 05:35 PM #1

- September 5th 2010, 08:00 AM #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

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.

- September 10th 2010, 03:18 AM #3