8. (10 points) As shown in the figure, in triangle $A B C$, $A F=2 B F$, $C E=3 A E$, $C D=4 B D$. Connect $C F$ and intersect $D E$ at

保留源文本的换行和格式，直接输出翻译结果如下：

8. (10 points) As shown in the figure, in triangle $A B C$, $A F=2 B F$, $C E=3 A E$, $C D=4 B D$. Connect $C F$ and intersect $D E$ at
【Analysis】Connect $E F, D F$, then according to the kite model, we have: $S_{\triangle E F C}: S_{\triangle F D C}=E P: P D$. We only need to find the ratio of $S_{\triangle E F C}$ and $S_{\triangle F D C}$ to determine $E P: P D$. From the given proportional relationships between line segments, we can calculate the ratio of the areas, and find the area ratios of $\triangle E F C$ and $\triangle F D C$ with respect to $\triangle A B C$, thus finally obtaining the value of $E P: P D$.

【Solution】Solution: According to the analysis, as shown in the figure, connect $E F, D F$, then according to the kite model, we have: $S_{\triangle E F C}: S_{\triangle F D C}=$ $E P: P D$.
Since $A F=2 B F$, we have $S_{\triangle A F C}: S_{\triangle B F C}=A F: B F=2: 1 \Rightarrow S_{\triangle \mathrm{AFC}}=\frac{2}{3} S_{\triangle \mathrm{ABC}}, S_{\triangle \mathrm{BFC}}=\frac{1}{3} S_{\triangle A B C}$; similarly, $C E=3 A E \Rightarrow S_{\triangle E F C}: S_{\triangle A E F}=E C: A E=3: 1 \Rightarrow S_{\triangle E F C}=\frac{3}{4} \times S_{\triangle A F C}=\frac{3}{4} \times \frac{2}{3} \times S_{\triangle A B C}$ $=\frac{1}{2} S_{\triangle A B C}$;
$C D=4 B D \Rightarrow S_{\triangle C D F}: S_{\triangle B D F}=C D: B D=4: 1 \Rightarrow S_{\mathrm{FDC}}=\frac{4}{5} \times S_{\triangle B C F}=\frac{4}{5} \times \frac{1}{3} \times S_{\triangle A B C}=$ $\frac{4}{15} \times S_{\triangle \mathrm{ABC}}$.
Therefore: $E P: P D=S_{\triangle E F C}: S_{\triangle F D C}=\frac{1}{2} S_{\triangle A B C}: \frac{4}{15} S_{\triangle A B C}=\frac{1}{2}: \frac{4}{15}=\frac{15}{8}$. Hence, the answer is: $\frac{15}{8}$.