## Task A-2.4.

Let $\overline{A B}$ and $\overline{C D}$ be diameters of the circle $k$ with center $S$ and let $\varangle B A D=28^{\circ}$. A circle with center $A$ passing through the point $S$ intersects the circle $k$ at points $E$ and $F$, with $D$ and $F$ on the same side of the line $A B$. Determine $\varangle C F S$.
## First solution.

Triangle $A F S$ is equilateral, so $\varangle A S F=60^{\circ}$. 1 point

Triangle $C S F$ is isosceles, so $\varangle C F S=\varangle S C F$. 1 point

Since angles $\angle D C F$ and $\angle D A F$ are inscribed angles subtended by the chord $\overline{D F}$, we have

$$
\varangle C F S=\varangle D C F=\varangle D A F .
$$

Finally, we conclude

$$
\varangle D A F=\varangle S A F-\varangle B A D=60^{\circ}-28^{\circ}=32^{\circ}
$$