[Sum of angles in a triangle. Theorem of the exterior angle.] [Median line of a triangle $]

On the side $AC$ of triangle $ABC$, a point $D$ is chosen such that $2AD = DC$. $E$ is the foot of the perpendicular dropped from point $D$ to segment $BC$, and $F$ is the intersection point of segments $BD$ and $AE$. Find the angle $ADB$, given that triangle $BEF$ is equilateral.
$\angle B D E=90^{\circ}-\angle D B E=90^{\circ}-60^{\circ}=30^{\circ}, \angle D E F=\angle B F E-\angle B D E=60^{\circ}-30^{\circ}=30^{\circ}$. Therefore, triangle $D F E$ is isosceles, $F D=F E=B F$. Hence, $F$ is the midpoint of segment $B D$. Let $K$ be the midpoint of $D C$. Then $K D=A D$, which means $D$ is the midpoint of $A K$. On the other hand, $F K$ is the midline of triangle $B D C$, so $F K \| B C$ and $\angle D F K=$ $\angle D B C=60^{\circ}=\angle A F D$.

This means that the median $F D$ of triangle $A F K$ is also its bisector. Therefore, triangle $A F K$ is isosceles, and $F D$ is its altitude, that is,

$\angle A D B=\angle A D F=90^{\circ}$.

## Answer

$90^{\circ}$.