Final Round

Let $K,L,N$ be points of intersection of the bisectors $AK,BL,CN$ and the circumcircle of a triangle $ABC$, respectively (see the Fig.). Let $I$ be the incenter of $△ABC$ and let the segment $NL$ meet $AC$ at $B_2^{\prime }$. Let $M$ be the intersection point of $CN$ and $AB$. By the trefoil theorem $AL=IL$, i.e. $△AIL$ is isosceles. Since $\angle ALN=\angle ACN=\angle NCB=\angle BLN$ (as inscribed angles subtending equal arcs), the ray $LN$ is the bisector of the isosceles triangle $AIL$, hence $LN$ is the perpendicular bisector of the side $AI$. Therefore, $△A{B}_2^{\prime }I$ is also isosceles and $\angle IA{B}_2^{\prime }=\angle B_2^{\prime }IA$. Since $AK$ is the bisector of the angle $BAC$, we have $\angle IA{B}_2^{\prime }=\angle BAI$, so $AB\parallel B_2^{\prime }I$. Then we have $\frac{C{B}_2^{\prime }}{B_2^{\prime }A}=[\text{the Thales theorem}]=\frac{CI}{MI}=\frac{AC}{AM}=$ $=\left[\frac{AM}{AB}=\frac{AC}{AC+BC}\right]=\frac{AC+BC}{AB}=\frac{a+b}{c}$, so $\frac{C{B}_2^{\prime }}{AC}=\frac{a+b}{a+b+c}$. Therefore $B_2^{\prime }$ coincide with $B_2$. Thus, $B_2$ and $C_1$ lie on the segment $NL$. Similarly, we see that $B_1$ and $A_2$ lie on $KL$, $A_1$ and $C_2$ lie on $KN$, which gives the required statement.