$17 \cdot 166$ An equilateral $\triangle D E F$ is inscribed in an equilateral $\triangle A B C$ (as shown in the figure), and $D E \perp B C$. Then $S_{\triangle D E F}: S_{\triangle A B C}$ equals
(A) $\frac{1}{6}$.
(B) $\frac{1}{4}$.
(C) $\frac{1}{3}$.
(D) $\frac{2}{5}$.
(E) $\frac{1}{2}$.
(46th American High School Mathematics Examination, 1995)
[Solution] Clearly, $\triangle A F E \cong \triangle B D F \cong \triangle C E D$, so $B D=E C$.
$\therefore \frac{C D}{B D}=\frac{C D}{E C}=\cos \angle C=\frac{1}{2}$, which means $C D=\frac{1}{3} B C$.
And, $E D=\sqrt{3} \quad C D=\frac{\sqrt{3}}{3} B C$, thus $\frac{D E}{B C}=\frac{\sqrt{3}}{3}$.
$\therefore \quad S_{\triangle D E F}: S_{\triangle A B C}=E D^{2}: B C^{2}=1: 3$.
Therefore, the answer is $(C)$.