17.175 As shown in the figure, the side $AB$ of $\triangle ABC$ is divided into $n$ equal parts $(n>1990)$, and through these $n-1$ division points, lines $B_{i} C_{i} / / B C \quad(i=1$, $2,3, \cdots, n-1)$ are drawn, then
$$
S_{\triangle A B_{1} C_{1}}: S_{\text {quadrilateral } B_{1989} B_{1990} C_{1990} C_{1989}}
$$

is
(A) $1: 1989$.
(B) $1: 1990$.
(C) 1:3979.
(D) $1: 3977$.
(6th "Jinyun Cup" Junior High School Mathematics Invitational Competition, 1989)
[Solution] From the given and the properties of similar figures, we have
$$
\frac{S_{\triangle A B_{1} C_{1}}}{S_{\triangle A B_{1990}} C_{1990}}=\frac{1}{1990^{2}},
$$

Also, \( S_{\text {quadrilateral } B_{1989} B_{1990} C_{1990} C_{1989}}=S_{\triangle A B_{1990} C_{1990}}-S_{\triangle A B_{1989} C_{1989}} \),
thus
$$
\begin{array}{l}
\therefore \quad \frac{S_{\triangle A B_{1} C_{1}}}{S_{\text {quadrilateral } B_{1989} B_{1990} C_{1990} C_{1989}}}=\frac{1}{1990^{2}-1989^{2}} \\
=\frac{1}{(1990+1989)(1990-1989)}=\frac{1}{3979} \text {. } \\
\end{array}
$$

Therefore, the answer is \((C)\).