\left.\begin{array}{l}{[\text { Angles subtending equal arcs and equal chords }]} \\ {[\text { Inscribed angle subtending the diameter }]\end{array}\right]

A circle passes through vertices $A$ and $C$ of triangle $ABC$, intersecting side $AB$ at point $E$ and side $BC$ at point $F$. Angle $AEC$ is five times the angle $BAF$, and angle $ABC$ is $72^{\circ}$. Find the radius of the circle if $AC=6$.
Let $\angle B A F=\alpha$. Then $\angle A E C=5 \alpha$. Inscribed angles $E C F$ and $E A F$ subtend the same arc, so $\angle B C E=\angle E C F=\angle E A F=\angle B A F=\alpha$.

By the exterior angle theorem of a triangle, $\angle A B C+\angle B C E=\angle A E C$, or $72^{\circ}+\alpha=5 \alpha$, from which we find that $\alpha=\frac{1}{4} \cdot 72^{\circ}=18^{\circ}$, hence $\angle A E C=5 \alpha=5 \cdot 18^{\circ}=90^{\circ}$.

The center of the circle circumscribed around the right triangle $A E C$ is the midpoint of the hypotenuse $A C$, so the radius of the circle is half of $A C$, i.e., 3.