Junior Balkan Mathematical Olympiad

Let the circles with diameter $EM$ and $FM$ intersect for second time at $D^{\prime }$ and let them intersect the sides $CA$, $CB$ at points $G,K$ respectively. Since $$\angle E{D}^{\prime }M=\angle F{D}^{\prime }M=90^{\circ },$$ we have that $E,D^{\prime },F$ 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 $C,D^{\prime },M$ are collinear. From the above it follows that $D^{\prime }\equiv D$. Finally, from the cyclic quadrilaterals $EAMD$ and $DMBF$ we have that $$\angle ADB=180^{\circ }-\angle EDA-\angle BDF=180^{\circ }-\angle AME-\angle BMF=\angle EMF.$$

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Alternative solution.

Let $H,G$ be the points of intersection of $ME,MF$ with $AC,BC$ respectively. From the similarity of triangles $△MHA$ and $△MAE$ we get $$\frac{MH}{MA}=\frac{MA}{ME},$$ thus, $$M{A}^2=MH\cdot ME.\text{\hspace{1em}(1)}$$ Similarly, from the similarity of triangles $△MBG$ and $△MFB$ we get $$\frac{MB}{MF}=\frac{MG}{MB}$$ thus, $$M{B}^2=MF\cdot MG.\text{\hspace{1em}(2)}$$ 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 $△EMF$ and $△ADB$ are similar, so $△ADB=\angle EMF$. Even more $\angle ADB=\angle ADM+\angle MDB=\angle AEM+\angle MFB=\angle CAB+\angle CBA$.