ABC is an acute-angled triangle with circumcenter O and orthocenter H (and O ≠ H). Show that one of area AOH, area BOH, area COH is the sum of the other two.
Solution
Let G be the centroid. Note that OH is the Euler line, so G lies on OH. wlog A is on one side of the line OH, and B, C are on the other side. Let M be the midpoint of BC. Then the length of the perpendicular from M to the line OH is the mean of the lengths of the perpendiculars from B and C. Thus if ∠MGO = θ, then area BOH + area COH = OH·GM sin θ = (OH·AG/2) sin θ = area AOH.
© John Scholes
jscholes@kalva.demon.co.uk
22 Mar 2004
Last corrected/updated 22 Mar 04