Belarusian Mathematical Olympiad

Let $O$ be the circumcenter of the triangle $△ABC$. Let $\angle BAC=\alpha$, $\angle ABC=\beta$, $\angle ACB=\gamma$. We construct the line passing through $D$ and $O$. Let $L$ be the point of intersection of this line and the ray $BI$. Since $AI$ is a bisector of the angle $BAC$, we have $BD=DC$, and so the line $DO$ is a perpendicular bisector of the segment $BC$. Therefore, $BL=CL$, i.e. the triangle $BLC$ is isosceles and $\angle LBC=\angle LCB=\beta /2$ ($BI$ is the bisector of the angle $ABC$). Hence,



$$\angle ILC=\angle BLC=180^{\circ }-(\angle LBC+\angle LCB)=180^{\circ }-\beta .$$

On the other hand, $$\begin{array}{rl}\angle CDI& =\angle CDA=180^{\circ }-(\angle DAC+\angle ACD)=\\ & =[\angle DCB=\angle DAB=\angle DAC=0.5\alpha ,\angle ACB=\gamma ,\angle ACD=\\ & =\angle ACB+\angle DCB=\gamma +0.5\alpha ]=180^{\circ }-(0.5\alpha +\gamma +0.5\alpha )=180^{\circ }-\alpha -\gamma =\beta .\end{array}$$ Therefore, $\angle ILC+\angle CDI=180^{\circ }$. It follows that the points $I$, $D$, $C$, and $L$ lie on the same circumference (passing through $I$, $D$, $C$). Thus, $K$ and $L$ coincide. Hence, $CK=CL=BL=BK$, as required.