O is the circumcenter of the acute-angled triangle ABC. The altitudes are AD, BE and CF. The line EF cuts the circumcircle at P and Q. Show that OA is perpendicular to PQ. If M is the midpoint of BC, show that AP2 = 2 AD·OM.
Solution
Let OA and PQ meet at T. ∠AEH = ∠AFH = 90o, so AEHF is cyclic, so ∠AFT = ∠AFE (same angle) = ∠AHE = 90o- ∠HAE = 90o - ∠DAC (same angle) = ∠C. But ∠TAF = ∠OAF (same angle) = 90o - (1/2) ∠AOB = 90o - ∠C. Hence ∠AFT = 90o, which establishes that OA and PQ are perpendicular.
Let the circumradius be R and let AA' be a diameter. We have AF = AC cos A = 2R sin B cos A. Hence AT = AF cos OAB = AF sin C = 2R cos A sin B sin C. Now PT2 = PT·TQ = AT.A'T = AT(2R - AT). Hence AP2 = 2R·AT = 4R2 cos A sin B sin C.
We have AD = AC sin C = 2R sin B sin C, and OM = OC cos COM = R cos A. Hence 2 AD·OM = AP2.
© John Scholes
jscholes@kalva.demon.co.uk
24 Oct 2002