IMO Shortlist, 2019

Denote the centres of $\omega_{B}$ and $\omega_{C}$ by $O_{B}$ and $O_{C}$, let their radii be $r_{B}$ and $r_{C}$, and let $BC$ be tangent to the two circles at $T$ and $U$, respectively.

From the cyclic quadrilaterals $AFDC$ and $ABDE$ we have
$$\angle MDO_{B} = \frac{1}{2} \angle FDB = \frac{1}{2} \angle BAC = \frac{1}{2} \angle CDE = \angle O_{C}DN,$$
so the right-angled triangles $DMO_{B}$ and $DNO_{C}$ are similar. The ratio of similarity between the two triangles is
$$\frac{DN}{DM} = \frac{O_{C}N}{O_{B}M} = \frac{r_{C}}{r_{B}}.$$
Let $\varphi = \angle DMN$ and $\psi = \angle MND$. The lines $FM$ and $EN$ are tangent to $\omega_{B}$ and $\omega_{C}$, respectively, so
$$\angle MTP = \angle FMP = \angle DMN = \varphi \quad \text{and} \quad \angle QUN = \angle QNE = \angle MND = \psi.$$
(It is possible that $P$ or $Q$ coincides with $T$ or $U$, or lie inside triangles $DMT$ or $DUN$, respectively. To reduce case-sensitivity, we may use directed angles or simply ignore angles $MTP$ and $QUN$.)

In the circles $\omega_{B}$ and $\omega_{C}$ the lengths of chords $MP$ and $NQ$ are
$$MP = 2r_{B} \cdot \sin \angle MTP = 2r_{B} \cdot \sin \varphi \quad \text{and} \quad NQ = 2r_{C} \cdot \sin \angle QUN = 2r_{C} \cdot \sin \psi.$$
By applying the sine rule to triangle $DNM$ we get
$$\frac{DN}{DM} = \frac{\sin \angle DMN}{\sin \angle MND} = \frac{\sin \varphi}{\sin \psi}.$$
Finally, putting the above observations together, we get
$$\frac{MP}{NQ} = \frac{2r_{B} \sin \varphi}{2r_{C} \sin \psi} = \frac{r_{B}}{r_{C}} \cdot \frac{\sin \varphi}{\sin \psi} = \frac{DM}{DN} \cdot \frac{\sin \varphi}{\sin \psi} = \frac{\sin \psi}{\sin \varphi} \cdot \frac{\sin \varphi}{\sin \psi} = 1,$$
so $MP = NQ$ as required.