Problem 1943

If $AB = AC$, then the statement is obvious. Without loss of generality, we assume that $AB < AC$. Let $AH$ be the common chord of the given circles, as shown in the figure. We draw the line which passes through $A$ and is perpendicular to $AD$ and denote by $K$ and $L$ the points of intersection of the given lines with $\omega_1$ and $\omega_2$ respectively.

We prove that $BL$ passes through $M$. Let $X$ be the point of intersection of $BL$ and $AD$. Since $KB \parallel AD \parallel LC$ we have the following proportions:
$$\frac{AX}{KB} = \frac{LA}{LK}, \quad \frac{DX}{CL} = \frac{BD}{BC}, \quad \frac{BD}{BC} = \frac{KA}{LK}.$$
Thus, $AX = \frac{KB \cdot LA}{LK}$, $DX = \frac{CL \cdot KA}{LK}$. Also, $\angle KAB = \angle LAC$ and triangles $AKB$ and $ALC$ are similar. Thus, $\frac{KA}{LA} = \frac{KB}{LC}$, $KA \cdot LC = KB \cdot LA$. Hence, $AX = DX$ and $X$ coincides with $M$. By analogy, we can show that $CK$ passes through $M$ too. We have
$$\angle DME = \angle LMA = \angle CLE = 180^\circ - \angle DHE.$$

This implies that $E, M, D, H$ belong to the same circle. $KBFH$ is inscribed in the circle $\omega_1$ and thus
$$\angle DMF = \angle KMA = \angle MKB = 180^\circ - \angle BHF = \angle DHF,$$
which implies that $M, H, D, F$ lie on the circle.

We have shown that $M, H, D, F, E$ lie on the same circle. In the right angled triangle $HAD$, $HM$ is a median. Therefore, $MD = MH$. Thus
$$\angle MDH = \angle MHD = \angle MED = \angle MFH.$$
Consider triangles $MDE$ and $MBD$ that have the common angle $M$ and
$$180^\circ - \angle CFE = \angle MFE = \angle MDE = \angle MBD.$$
We have that $B, E, F, C$ are concyclic, which implies the result.