Example 1. As shown in the figure, $B D$: $D C$: $C E$: $E A=2$ : $1, A D, B E$ intersect at $F$, then $A F: F D=$ $\qquad$
$$
\begin{array}{l}
\because \lambda_{1}=\lambda_{2}=\frac{1}{2}, \\
\therefore \lambda_{3}=\left(1+\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{3}{4} .
\end{array}
$$

In this problem, $\triangle ABC$ only has the characteristics of a basic figure, and $D, F$ are known fixed points dividing $BC, AC$ respectively. Therefore, we can directly use (*) to solve.
$$
\begin{array}{l}
\because \lambda_{1}=\lambda_{2}=\frac{1}{2}, \\
\therefore \lambda_{3}=\left(1+\frac{1}{2}\right) \cdot \frac{1}{2}=\frac{3}{4} .
\end{array}
$$