$17 \cdot 169$ On the sides $AB, BC, CA$ of $\triangle ABC$, points $D, E, F$ are taken such that $AD:DB = BE:CE = CF:FA = 1:n$. Then the ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is
(A) $\frac{n^{2}-n+1}{(n+1)^{2}}$.
(B) $\frac{1}{(n+1)^{2}}$.
(C) $\frac{2 n^{2}}{(n+1)^{2}}$.
(D) $\frac{n^{3}}{(n+1)^{3}}$.
(E) $\frac{n(n-1)}{n+1}$.
(18th American High School Mathematics Examination, 1967)
[Solution] Let the three sides of $\triangle ABC$ be $a, b, c$, and the heights on these sides be $h_{a}, h_{b}, h_{c}$, respectively.
Let $S = S_{\triangle ABC}$, and let the heights on sides $AD, BE, CF$ of $\triangle ADF, \triangle BED, \triangle CEF$ be $l_{c}, l_{a}, l_{b}$, respectively. By the given conditions,
$$
\frac{l_{a}}{h_{a}}=\frac{l_{b}}{h_{b}}=\frac{l_{c}}{h_{c}}=\frac{n}{n+1},
$$

Thus, $S_{\triangle EFD} = S_{\triangle ABC} - S_{\triangle ADF} - S_{\triangle BED} - S_{\triangle CEF}$
$$
\begin{array}{l}
= S - \frac{n}{(n+1)^{2}}\left(\frac{1}{2} c h_{c} + \frac{1}{2} b h_{b} + \frac{1}{2} a h_{a}\right) \\
= S - \frac{3 n S}{(n+1)^{2}} = \frac{n^{2} - n + 1}{(n+1)^{2}} S .
\end{array}
$$

Therefore, the answer is $(A)$.