Example 8. In $\triangle ABC$, $D$ is the midpoint of $AC$.
(1) If $BC$ is trisected, with division points from $C$ being $C_{1}, C_{2}$, and $C_{1}A, C_{2}A$ intersect $BD$ at $D_{1}$, $D_{2}$. Find $S \wedge N_{1} D_{2}: S \triangle A C_{1} C_{2}$;
(2) If $BC$ is divided into four equal parts, with division points from $C$ being $C_{1}, C_{2}, C_{3}$, and $C_{1}A, C_{2}A$ intersect $BD$ at $D_{1}, D_{2}$. Find $S \triangle A D_{1} D_{2}: S \triangle A C_{1} C_{2}$;
(3) If $BC$ is divided into $n$ equal parts, with division points from $C$ being $C_{1}, C_{2}, \cdots, C_{n-1}$, and $C_{1}A, C_{2}A$ intersect $BD$ at $D_{1}, D_{2}$, then $S \triangle A D_{1} D_{2}: S \triangle A C_{1} C_{2}=$ ?
To solve, we only need (3).
It is easy to see that $B C_{2}={ }_{n}^{n \cdots 2} B C, B C_{1}=\frac{n-1}{n} B C$.
Let $\angle A B D=\alpha$,
$\angle D B C=\beta,$
Since $D$ is the point dividing $A C$ internally, by the angle bisector theorem, we can get
$\sin \alpha / \sin \beta=$
$B C / A B$, thus
$$
\begin{array}{l}
A D_{2}=\frac{A B \cdot \sin C}{B C_{2} \cdot \sin ?}=\frac{A B}{B C_{2}} \cdot \frac{B C}{A B} \\
D_{2} C_{2} \\
=\frac{B C}{B C_{2}}=\frac{n}{n-2}
\end{array}
$$

Similarly, we can get $A D_{1}^{1}=\frac{n}{2 n-1}$.
Therefore, if we denote $\angle C_{1} A C_{2}=\theta$, we immediately get
$$
\begin{array}{l}
\left.S_{1} \rightarrow 11\right) \frac{1}{2} A D_{1} \cdot A D_{2} \cdot \sin \theta \\
S \triangle A_{1} 1 C_{2} \frac{1}{2} A C_{1} \cdot A C_{2} \cdot \sin \pi \\
-\frac{A D}{A C_{1}} \cdot \frac{A D}{A C_{1}} \text {. } \\
=\frac{n^{2}}{2(n-1)(2 n-1)} \text {. } \\
\because n=3 \text {, } S \triangle \mathrm{AD}_{1} \mathrm{D}_{2}: S \triangle \triangle_{C_{1} C_{2}}=9: 20 \text {, } \\
\because n=4 \text {, } S \triangle \triangle_{\mathrm{D}_{1} D_{2}}: S \triangle \mathrm{AC}_{1} \mathrm{C}:=8: 21 \text {. } \\
\end{array}
$$