# Thread: chaleenging one

1. ## 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

2. 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.

3. yea,thanks got it