In $\triangle A B C$, extend $A B$, $A C$, and $B C$ successively, such that $B A^{\prime}=\lambda_{1} A B$, $C B^{\prime}=\lambda_{2} B C$, $A C^{\prime}=\lambda_{3} C A$. Find $S \triangle \triangle^{\prime} \mathrm{B}^{\prime} C^{\prime}: S \triangle \triangle B C$.
Solve for
$$
\begin{array}{l}
\frac{S \triangle A^{\prime} C^{\prime}}{S \triangle A B C}=\frac{A A^{\prime} \cdot A C^{\prime}}{A B \cdot A C} \\
\quad=\frac{\left(1+\lambda_{1}\right) A B \cdot \lambda_{3} C A}{A B \cdot A C} . \\
=\left(1+\lambda_{1}\right) \lambda_{3},
\end{array}
$$

Thus, $S \triangle A^{\prime} C^{\prime}=\left(1+\lambda_{1}\right) \lambda_{3} S \triangle A B C$.
$$
S \triangle C^{\prime} C B^{\prime}=\left|\left(\lambda_{3}-1\right) \lambda_{2}\right| S \triangle A B C \text {. }
$$

Also, $S \triangle A^{\prime} B^{\prime} C^{\prime}=S \triangle A^{\prime} A C^{\prime}+S \triangle C^{\prime} C B^{\prime}$
$-S \triangle A B C-S \triangle A^{\prime} B B^{\prime}$,

Thus, $\frac{S \triangle A^{\prime} B^{\prime} C^{\prime}}{S \triangle A B C}=1+\left(\lambda_{1}+\lambda_{2}-\lambda_{3}\right)$
$$
+\left(\lambda_{1} \lambda_{2}-\lambda_{2} \lambda_{3}-\lambda_{3} \lambda_{1}\right) \mid \text {. }
$$

When $\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda$, we have
$S \triangle A B C$