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 $BB_1$. Thus, $NA = NB = NB_1$, 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 $QB_1$ contains the symmedian of triangle $ABB_1$ 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
$$\angle(B_1C, B_1E) = \angle(B_1C, B_1Q) = \angle(B_1M, B_1B) = \angle(PM, PB) = \alpha. \qquad (1)$$
Also, the following equalities hold:
$$\angle(CK, CA) = \angle(BK, BA) = \angle(BK, BM) = \angle(PK, PM) = \beta. \qquad (2)$$
From (1) and (2) it follows that $\angle(PK, B_1E) = \angle(CK, B_1E) = \alpha + \beta = \angle(PK, PB)$. The latter equality gives
$$B_1E \parallel PB. \qquad (3)$$
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 $B_1E$, since (3) holds.