SILK ROAD MATHEMATICS COMPETITION XX

Let the bisector of $\angle ACB$ intersect $\omega$ for the second time at point $N$. Note that $N$ is the midpoint of arc $AB$ (that does not contain $C$) of $\omega$. Also, it is easy to see that line $CN$ is the perpendicular bisector of segment $B{B}_1$. Thus, $NA=NB=N{B}_1$, i.e. points $A,B$ and $B_1$ lie on a circle centered at $N$ with radius $NA$. Since $\angle NAQ=\angle NBQ=90^{\circ }$, then lines $QA$ and $QB$ are tangent to this circle. Therefore, line $Q{B}_1$ contains the symmedian of triangle $AB{B}_1$ corresponding to vertex $B_1$. Let line $KC$ intersect ${\omega }_1$ for the second time at point $P$. From the properties of symmedian it follows that $$\angle (B_{1}C,B_{1}E)=\angle (B_{1}C,B_{1}Q)=\angle (B_{1}M,B_{1}B)=\angle (PM,PB)=\alpha .(1)$$ Also, the following equalities hold: $$\angle (CK,CA)=\angle (BK,BA)=\angle (BK,BM)=\angle (PK,PM)=\beta .(2)$$ From (1) and (2) it follows that $\angle (PK,B_{1}E)=\angle (CK,B_{1}E)=\alpha +\beta =\angle (PK,PB)$. The latter equality gives $$B_{1}E\parallel PB.(3)$$ Let line $BP$ intersect line $AC$ at point $R$. Then, from (2) it follows that $CA\parallel PM$, or that $MP$ is a midsegment of triangle $ARB$, i.e. $P$ is the midpoint of segment $BR$. Now it is enough to note that if line $KC$ bisects segment $BR$, then it also bisects segment $B_{1}E$, since (3) holds.