In ` A B C ,` the three bisectors of the angle A, B and C are extended to intersect the circumcircle at D,E and F respectively. Prove that `A DcosA/2+B EcosB/2+C FcosC/2=2R(sinA+sinB+sinC)`

Explanation

Using Sine law in `Delta ABD`,
we get

`(AD)/(sin(B + (A)/(2))) = 2R`
`:. AD = 2R sin (B + (A)/(2))`
`rArr AD cos.(A)/(2) = 2R cos.(A)/(2) sin (B + (A)/(2))`
`=R (sin (A + B) + sin B)`
`=R(sin C + sin B)`
Similarly, `BE cos.(B)/(2) = R (sin A + sin C)`
And `CF cos.(C)/(2) = R (sin A + sin B)`
`:. AD cos.(A)/(2) + BE cos.(B)/(2) + CF cos.(C)/(2)`
`=2R(sin A + sin B + sinC)`