58th Ukrainian National Mathematical Olympiad

(Anton Trigub) Fig. 46 Suppose that lines $AI$ and $CI$ meet the circumcircle of $△ABC$ (second time) in the points $W_A$ and $W_C$, respectively (Fig. 46). Since $\angle AKI=\angle AIC$, then circumcircle of the triangle $AKI$ tangent to the line $W_{C}I$. Consider the triangle $A{W}_{C}I$. According to well-known fact this triangle is isosceles. Then the circumcircle of the triangle $△AKI$ tangents to $W_{C}A$ as well. Thus a symmedian of the $△AKI$ belongs to the line $W_{C}K$. Then $\angle W_{C}KI=180^{\circ }-\angle ICA=180^{\circ }-\angle W_C{W}_{A}I$. It means that $K$ belongs to the circumcircle of the $△W_C{W}_{A}I$. In the same way $L$ belongs to this circumcircle as well. Let us note that $△W_{C}B{W}_A=△W_{C}I{W}_A$ (they have equal sides), and thus the radius of the circumcircle of $△KIL$ equal to the radius of the circumcircle of $△ABC$.