Question

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

Answer 1

Answer:

Hence proved that (r1/bc)+(r2/ca)+(r3/ab)=1/r-1/2R

Step-by-step explanation:

r1/bc + r2/ca + r3/ab = 1/abc[ar1 + br2 + cr3]

= 1/abcΣar1 = 1/abcΣa s tanA/2

= s/abc Σ2RsinAtanA/2

= s4R/abcΣsinA/2cosA/2Sin/Cos(A/2)/(A/2)

= s 1/Δ Σsin^2A/2

= 1/r Σ 1 - cos A/2

= 1/2r [ 1 - cosA + 1 - cosB + 1 - cosC ]

= 1/2r [ 3 - ( cos A + cos B + cos C)]

= 1/2r [ 3 - ( 1+ 4sinA/2sinB/2sinC/2)]

= 1/2r [ 2 -4RsinA/2sinB/2sinc/2/R]

= 1/2r [ 2 - r/R]

= 1/r - 1/2R

Hence proved