ABC is a triangle with incenter I. Show that the circumcenters of IAB, IBC, ICA lie on a circle whose center is the circumcenter of ABC.
Solution
In fact they lie on the circumcircle of ABC.
Extend AI to meet the circumcircle again at A'. We show that A' is the circumcenter of BCI. Angle A'AC = angle A'AB, so A' is the midpoint of the arc BC, so A'B = A'C. Also ∠A'CB = ∠A'AB = A/2, so ∠A'CI = A/2 + B/2. But ∠A'IC = ∠IAC + ∠ICA = A/2 + B/2, so A'CI is isosceles, so A'C = A'I.
© John Scholes
jscholes@kalva.demon.co.uk
21 Aug 2002