Junior Balkan Mathematical Olympiad

Let the circles with diameter $EM$ and $FM$ intersect for second time at $D'$ and let them intersect the sides $CA$, $CB$ at points $G,K$ respectively. Since
$$\angle ED'M = \angle FD'M = 90^{\circ},$$
we have that $E,D',F$ are collinear.
Since $EM$ is a diameter and $AG$ is a chord perpendicular to it, we have that $MG = MA$ and similarly $MK = MB$. Since $MA = MB$, it follows that $AGKB$ is cyclic.
From the above we have that $CG \cdot CA = CK \cdot CB$ and this means that $C$ has equal power to the two circles, so it is on the radical axis of them, so $C,D',M$ are collinear. From the above it follows that $D' \equiv D$.
Finally, from the cyclic quadrilaterals $EAMD$ and $DMBF$ we have that
$$\angle ADB = 180^{\circ} - \angle EDA - \angle BDF = 180^{\circ} - \angle AME - \angle BMF = \angle EMF.$$

Let $H,G$ be the points of intersection of $ME,MF$ with $AC,BC$ respectively. From the similarity of triangles $\triangle MHA$ and $\triangle MAE$ we get
$$\frac{MH}{MA} = \frac{MA}{ME},$$
thus,
$$MA^2 = MH \cdot ME. \tag{1}$$
Similarly, from the similarity of triangles $\triangle MBG$ and $\triangle MFB$ we get
$$\frac{MB}{MF} = \frac{MG}{MB}$$
thus,
$$MB^2 = MF \cdot MG. \tag{2}$$
Since $MA = MB$, from (1) and (2) we have that the points $E,H,G,F$ are concyclic.
Therefore, we get that $\angle FEH = \angle FEM = \angle HGM$. Also, the quadrilateral $CHMG$ is cyclic, so $\angle CMH = \angle HGC$. We have
$$\angle FEH + \angle CMH = \angle HGM + \angle HGC = 90^{\circ}.$$
Thus $CM \perp EF$. Now, from the cyclic quadrilaterals $FDMB$ and $EDMA$, we get that $\angle DFM = \angle DBM$ and $\angle DEM = \angle DAM$. Therefore, the triangles $\triangle EMF$ and $\triangle ADB$ are similar, so $\triangle ADB = \angle EMF$. Even more $\angle ADB = \angle ADM + \angle MDB = \angle AEM + \angle MFB = \angle CAB + \angle CBA$.