O is the circumcenter of the triangle ABC. The lines AO, BO, CO meet the opposite sides at D, E, F respectively. Show that 1/AD + 1/BE + 1/CF = 2/AO.
Solution
Projecting onto the altitude from A, we have AD cos(C - B) = AC sin C = 2R sin B sin C, so 2R/AD = cos(C - B)/(sin B sin C).
Hence 2R/AD + 2R/BE + 2R/CF =cos(C - B)/(sin B sin C) + cos(A - C)/(sin C sin A) + cos(B - A)/(sin A sin B). So 2R sin A sin B sin C (1/AD + 1/BE + 1/CF) = sin A cos(B - C) + sin B cos(C - A) + sin C cos(A - B) = 3 sin A sin B sin C + sin A cos B cos C + sin B cos A cos C + sin C cos A cos B = 3 sin A sin B sin C + sin(A + B) cos C + sin C cos A cos B = 3 sin A sin B sin C + sin C (cos C + cos A cos B) = 3 sin A sin B sin C + sin C (-cos(A + B) + cos A cos B) = 4 sin A sin B sin C. Hence 1/AD + 1/BE + 1/CF = 2/R.
© John Scholes
jscholes@kalva.demon.co.uk
1 July 2002
Last corrected/updated 22 Oct 2002