20.71 As shown in the figure, $\triangle ABC$ is inscribed in a circle. Point $D$ is on $\overparen{AC}$, and $\overparen{DC} \stackrel{\mathrm{m}}{=} 30^{\circ}$. Point $G$ is on $\overparen{BA}$, and $\overparen{BG} > \overparen{GA}$. The lengths of sides $AB$ and $AC$ are equal to the length of chord $DG$, and $\angle CAB = 30^{\circ}$. Chord $DG$ intersects sides $AC$ and $AB$ at points $E$ and $F$, respectively. The ratio of the area of $\triangle AFE$ to the area of $\triangle ABC$ is
(A) $\frac{2-\sqrt{3}}{3}$.
(B) $\frac{2 \sqrt{3}-3}{3}$.
(C) $7 \sqrt{3}-12$.
(D) $3 \sqrt{3}-5$.
(E) $\frac{9-5 \sqrt{3}}{3}$.
(32nd American High School Mathematics Examination, 1981)
[Solution] As shown in the figure, connect $DC$. Since $AC \stackrel{\mathrm{m}}{=} 2 \angle ABC$, then by $\angle ABC = (180^{\circ} - 30^{\circ}) / 2 = 75^{\circ}$, so $\overparen{AC} \stackrel{\mathrm{m}}{=} 150^{\circ}$.
Thus, $\overparen{AD} = \overparen{AC} - \overparen{DC} \stackrel{\mathrm{m}}{=} 150^{\circ} - 30^{\circ} = 120^{\circ}$,
so $\angle ACD = 60^{\circ}$.
Since $AC = DG$, then $\overparen{GA} = \overparen{GD} - \overparen{AD} \stackrel{\mathrm{m}}{=} \overparen{AC} - 120^{\circ} = 30^{\circ}$,
and $\overparen{CG} = \overparen{AC} + \overparen{GA} \stackrel{\mathrm{m}}{=} 150^{\circ} + 30^{\circ} = 180^{\circ}$, and $\angle CDG = 90^{\circ}$.
Therefore, $\triangle DEC$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ right triangle.
Without loss of generality, assume $AC = AB = DG = 1, AE = x = DE$ (it is easy to prove $\triangle EDC \cong \triangle EAG$, so $AE = DE$).
$\therefore \quad CE = 1 - x = \frac{x}{\sqrt{3}} \cdot 2$.
Then $AE = \dot{x} = 2\sqrt{3} - 3$.
$FH$ is the altitude from $A$ to $AE$ in $\triangle AFE$, so
$$
\begin{array}{c}
EH = \frac{AE}{2} = \frac{2\sqrt{3} - 3}{2}, \text{ and } \\
FH = \frac{2\sqrt{3} - 3}{2} \cdot \frac{\sqrt{3}}{3}. \\
\therefore \quad S_{\triangle AFE} = EH \cdot FH = \left(\frac{2\sqrt{3} - 3}{2}\right)^2 \cdot \frac{\sqrt{3}}{3} \\
= \frac{7\sqrt{3} - 12}{4}.
\end{array}
$$

And $S_{\triangle ABC} = \frac{1}{2} AB \cdot AC \sin 30^{\circ} = \frac{1}{2} \cdot 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{4}$.
$$
\therefore \quad \frac{S_{\triangle AEF}}{S_{\triangle ABC}} = 7\sqrt{3} - 12.
$$

Therefore, the answer is $(C)$.