$\left[\begin{array}{l}\text { Symmetry helps solve the problem } \\ \text { [The inscribed angle is half the central angle]}\end{array}\right]$

$A B$ is the diameter of the circle; $C, D, E$ are points on the same semicircle $A C D E B$. On the diameter $A B$, points $F$ and $G$ are taken such that $\angle C F A = \angle D F B$ and $\angle D G A = \angle E G B$. Find $\angle F D G$, if the arc $A C$ is $60^{\circ}$ and the arc $B E$ is $20^{\circ}$.
Let $C_{1}$ and $E_{1}$ be the points of intersection of the rays $D F$ and $D G$ with the given circle. Then $\cup A C_{1}=\cup A C$ and $\cup$ $B E_{1}=\cup B E$.

## Solution

Let $C_{1}$ and $E_{1}$ be the points of intersection of the rays $D F$ and $D G$ with the given circle. Since the circle is symmetric with respect to the diameter $A B$,
the point $C_{1}$ is symmetric to the point $C$ with respect to $A B$. Therefore, $\cup A C_{1}=\cup A C$. Similarly, $\cup B E_{1}=\cup B E$. Consequently,

$$
\angle C_{1} D E_{1}=\frac{1}{2} \cup C_{1} E_{1}=\frac{1}{2}\left(180^{\circ}-60^{\circ}-20^{\circ}\right)=50^{\circ}
$$

## Answer

$50^{\circ}$.