Problem 984

Solution. The triangles $A D F$ and $D B E$ are similar to the triangle $A B C$. Using the usual notation, $\angle A F D = \angle D E B = \gamma$. Therefore, the tangents to the circles $A D F$ and $D E B$ at point $D$ form an angle $\gamma$ with the line $A B$. Since $\gamma < 90^{\circ}$, these circles intersect on the side of the line $A B$ that contains $C$, so the point $G$ also lies on this side of the line $A B$.

The quadrilateral $A B E F$ is a cyclic quadrilateral if and only if $\angle C E F = \alpha$ and $\angle C F E = \beta$ (these two conditions are equivalent to each other), which means that the line $E F$ is a common tangent to the circles $A D F$ and $D B E$.

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In this case, the point $G$ clearly lies inside the triangle $E F D$. We can assume that the point $G$ is an interior point of the triangle $A B C$, otherwise, $A B E F$ cannot be a cyclic quadrilateral, and $G$ cannot lie on the segment $C D$. Based on the diagram, then $\angle E G D = 180^{\circ} - \beta$ and $\angle F G D = 180^{\circ} - \alpha$, it is evident that $\angle E G F = 180^{\circ} - \gamma$, so the quadrilateral $E C F G$ is a cyclic quadrilateral.

In summary: the quadrilateral $A B E F$ is a cyclic quadrilateral if and only if $\angle C F E = \beta$, or if $\angle C G E = \beta$. Considering that $\angle E G D = 180^{\circ} - \beta$, this is equivalent to the angle $\angle C G D$ being a straight angle, or that the point $G$ lies on the segment $C D$.