The diameters of the circumcirle of triangle ABC drawn from A,B and C meet BC, CA and AB, respectively, in L,M and N. Prove that `(1)/(AL) + (1)/(BM) + (1)/(CN) = (2)/(R)`

Explanation

Using sine rule in `DeltaOBL`, we get
`(R)/(sin[(pi)/(2) + C - B]) = (OL)/(sin ((pi)/(2) -A))`
`:. OL = (R cos A)/(cos(C - B))`
`rArr AL = R + (R cos A)/(cos (C - B))`
`= R (cos (C -B) + cos A)/(cos(C -B))`
`=R(cos (C -B) - cos (B + C))/(cos (C -B))`
`= (2R sin B sin C)/(cos (C - B))`
`rArr (1)/(AL) = (cos (C - B))/(2R sin B sin C) = (2 cos (C - B) sin A)/(4R sin B sin C sin A) = (sin 2B + sin 2C)/(4R sin B sin C sin A)`
`rArr (1)/(AL) + (1)/(BM) + (1)/(CN)`
`= (2(sin 2A + sin 2B + sin 2C))/(4R sin A sin B sin C)`
`= (2)/(R)`