Let $\triangle A B C$ have an area of $1,0<x \leqslant 1, A^{\prime}, B^{\prime}, C^{\prime}$ be points on $B C, C A, A B$ respectively, and $B A^{\prime}: A^{\prime} C=C B^{\prime}: B^{\prime} A$ $\Rightarrow A C^{\prime}=C^{\prime} B=(1-x)+x$. Try to express the area of $\triangle A^{\prime} B^{\prime} C^{\prime}$ using $x$.
$\cdots$ 、Obviously, $B A^{\prime}=(1-x) B C, B C^{\prime}=x B A$, therefore
$$
\begin{array}{l}
S_{\triangle C^{\prime}}=\frac{1}{2} B A^{\prime} \cdot B C^{\prime} \sin B \\
=\frac{1}{2}(1-x) x \\
=B A \cdot B C \sin B \\
=x(1-x) \\
\quad \cdot S_{\triangle A B C} \\
=x(1-x) .
\end{array}
$$

I can still deduce
$$
\begin{array}{l}
S_{\triangle A^{\prime} E^{\prime}} S_{\triangle B^{\prime} A}=x(1-x) \\
S_{\triangle A^{\prime} X^{\prime}}=S_{\triangle A B C}-S_{\triangle C B A^{\prime}}-S_{\triangle A^{\prime} C B^{\prime}}-S_{\triangle B A C^{\prime}} \\
=3 x^{2}-3 x+1
\end{array}
$$