Let the line $BQ$ meet the circle $\omega_1$ again at the point $T$. We will show that the points $M$, $T$, $N$ are collinear. Since $OQ$ is the diameter of the circle $\omega_1$, $MQ$ and $CQ$ are tangents to the circle $\omega$. Let $\angle ABC = \beta$. Then $\angle CAQ = \angle ACQ = \beta$ (as angles between tangents and a chord), $\angle ATQ = \angle ACQ = \beta$ and $\angle CTQ = \angle CAQ = \beta$ (as inscribed angles in the circle $\omega_1$), $\angle BNA = \angle AMB = \angle CAQ = \beta$ (as angles at parallel lines). Hence $\angle AMB + \angle ATB = \beta + (180^\circ - \beta) = 180^\circ$, so the quadrilateral $AMBT$ is cyclic and $\angle BTM = \angle MAB = \angle ACB$.

Next, depending on the order of the points $A$, $C$, and $N$ on the line, we consider the following cases. If the point $C$ lies between $A$ and $N$, then $\angle BNC + \angle BTC = \beta + (180^\circ - \beta) = 180^\circ$. Thus, the quadrilateral $BNCT$ is cyclic, and $\angle BTN = \angle BCN = 180^\circ - \angle ACB = 180^\circ - \angle BTM$. If the point $N$ lies between $A$ and $C$, then $\angle BNC = 180^\circ - \beta = \angle BTC$. Then the quadrilateral $BCNT$ is cyclic, and $\angle BTN = 180^\circ - \angle ACB = 180^\circ - \angle BTM$. We see that in both cases, the equality $\angle BTM + \angle BTN = 180^\circ$ holds, which proves that $T$ is the intersection point of the lines $MN$ and $BQ$.

Another solution, proposed by 11th grade student Roman Cheplyaka (Odessa), is based on an additional construction. Namely, construct the circle $\omega_2$ on the segment $OB$ as a diameter. Let $T$ be the point of intersection of the circles $\omega_1$ and $\omega_2$ different from $O$. Since $\angle OTB = \angle OTQ = 90^\circ$ as inscribed angles subtended by the diameter, the point $T$ lies on $BQ$ (so this is the same point $T$ as in the first solution). Let $R$ be the midpoint of the side $AB$ (and the intersection point of the diagonals of the parallelogram $AMBN$). Then the line $OR$ is the perpendicular bisector of $AB$. Hence, $\angle ORB = 90^\circ$, and the point $R$ also lies on $\omega_2$. Since the inscribed angles $RTB$ and $BOR$ in the circle $\omega_2$ are equal, and $\angle BOR = \frac{1}{2}\angle AOB = \angle ACB$, we have $\angle BTR = \angle ACB$. In addition, $\angle AQT = \angle ACT$ as angles inscribed in $\omega_1$. Let $P$ be the intersection point of the lines $AC$ and $RT$. Since the angle $CAR$ is acute and the angle $ART$ is right, these lines do intersect, and one of two cases occurs: either the point $P$ lies between $A$ and $C$, or the point $C$ lies between $A$ and $R$.

In the first case, we have the equalities $\angle BTP = 180^\circ - \angle BTR = 180^\circ - \angle BCP$, so the quadrilateral $BTPC$ is cyclic and $\angle ACT = \angle PCT = \angle PBT$. In the second case, $\angle BTP = 180^\circ - \angle BTR = 180^\circ - \angle ACB = \angle BCP$, so the quadrilateral $BTCP$ is cyclic and $\angle ACT = 180^\circ - \angle PCT = \angle PBT$. We see that in both cases the equality $\angle PBT = \angle ACT = \angle AQT$ holds, from which it follows that $BP \parallel AQ$. Thus, the point $P$ coincides with the point $N$, and the line $PR$ coincides with the line $MN$ (which contains the diagonal of the parallelogram $AMBN$). Therefore, the point $T$ lies simultaneously on the circle $\omega_1$ and on the lines $BQ$ and $MN$.