r1/bc+r2/ca+r3/ab=1/r-1/2R

r1/bc+r2/ca+r3/ab=1/r-1/2R

Topic: properties of triangle


Question:
r1/bc+r2/ca+r3/ab=1/r-1/2R

Answer:

given that 
r1/bc+r2/ca+r3/ab = 1/r-1/2R
LHS:
r1/bc+r2/ca+r3/ab
1⁄abc(𝝨a r1)
[r1=s. Tan(A/2)
a=2S sin (A)]
4RS/4R△ 𝝨sin2
=S/Δ 𝝨sin2A/2
1/r(sin2A/2+sin2B/2+sin2C/2)
1/r((1-cosA)/2+(1-cosB/2)/2+(1-cosC)/2)

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