Problem

Source: North Korea Team Selection Test 2013 #1

Tags: geometry, circumcircle, trigonometry, North Korea



The incircle of a non-isosceles triangle $ABC$ with the center $I$ touches the sides $ BC, CA, AB$ at $ A_1 , B_1 , C_1 $ respectively. The line $AI$ meets the circumcircle of $ABC$ at $A_2 $. The line $B_1 C_1 $ meets the line $BC$ at $A_3 $ and the line $A_2 A_3 $ meets the circumcircle of $ABC$ at $A_4 (\ne A_2 ) $. Define $B_4 , C_4 $ similarly. Prove that the lines $ AA_4 , BB_4 , CC_4 $ are concurrent.