The feet of the altitudes of the triangle ABC are D, E, F. The incircle of DEF meets its sides at G, H, I. Prove that ABC and GHI have the same Euler line (the line through the circumcenter and the centroid).
Solution
Let O be the intersection of the altitudes of ABC. ∠OEC = ∠ODC = 90o, so OECD is cyclic, so ∠ODE = ∠OCE. ∠ADC = ∠AFC = 90o, so AFDC is cyclic, so ∠OCE = ∠FCA = ∠FDA. So AD bisects ∠EDF. Hence O is the incenter of DEF. So it is the circumcenter of GHI.
Let H lie on DE and I on DF. Then DH and DI are tangents to the circle center O through H and I, so HI is perpendicular to DO. But so is BC, so HI is parallel to BC. Similarly GH is parallel to AB and IG to CA. So the Euler lines of GHI and ABC are parallel. But they both pass through O (which is the circumcenter of GHI and the orthocenter of ABC). Hence they coincide.
© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002