In $\triangle A B C$, points $D, E, F$ are on sides $B C, C A, A B$ respectively, and $A D, B E, C F$ intersect at point $P$. It is known that $\triangle A B C$ is divided into six smaller triangles by $A D, B E, C F$, and the areas of three of these smaller triangles are $6, 4, 5$ respectively. Find the values of the areas $x, y, z$ of the other three smaller triangles.
Solution: Notice that
$$
\begin{array}{l}
1=\frac{S_{\triangle A B P}}{S_{\triangle B D P}} \cdot \frac{S_{\triangle B D P}}{S_{\triangle B C P}} \cdot \frac{S_{\triangle B C P}}{S_{\triangle A B P}}=\frac{A P}{P D} \cdot \frac{D B}{B C} \cdot \frac{C E}{E A} \\
=\frac{S_{\triangle A P}}{S_{\triangle C P D}} \cdot \frac{S_{\triangle B B D}}{S_{\triangle B B C}} \cdot \frac{S_{\triangle P C E}}{S_{\triangle P E 1}}=\frac{10}{5} \cdot \frac{x}{x+5} \cdot \frac{4}{6} .
\end{array}
$$

Solving for $x$ yields $x=15$.
Since $A D, B E, C F$ are concurrent, we have
$$
\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=\frac{S_{\triangle A B P}}{S_{\triangle A C P}} \cdot \frac{S_{\triangle B C P}}{S_{\triangle A B P}} \cdot \frac{S_{\triangle A C P}}{S_{\triangle B C P}}=1 .
$$

Thus, $\frac{S_{\triangle P B D}}{S_{\triangle P D C}} \cdot \frac{S_{\triangle P C E}}{S_{\triangle P E A}} \cdot \frac{S_{\triangle P A F}}{S_{\triangle P F B}}=1$, which means
$$
\frac{x}{5} \cdot \frac{4}{6} \cdot \frac{z}{y}=1 \text {. }
$$

Substituting $x=15$ gives $y=2z$.
$$
\text { Also, } \frac{x}{5}=\frac{S_{\triangle P B D}}{S_{\triangle P D C}}=\frac{B D}{D C}=\frac{S_{\triangle \triangle B D}}{S_{\triangle A D C}}=\frac{x+y+z}{6+4+5} \text {, }
$$
Substituting $x=15$ gives $y+z=30$.
Therefore, solving yields $y=20, z=10$.
Hence, $S_{\triangle P B D}=x=15, S_{\triangle P F B}=y=20, S_{\triangle P A F}=z=10$.
(Huang Quanfu, Jiangzhen Middle School, Huaining County, Anhui Province, 246142)