12. As shown in the right figure, in triangle $A B C$, $D$ is the midpoint of $B C$, $A F=2 B F$, $C E=3 A E$. Connecting $C F$ intersects $D E$ at point $P$, find the value of $\frac{E P}{D P}$.
【Key Point】Geometry
【Answer】 3
【Analysis】Connect $E F, D F$
$$
\frac{E P}{D P}=\frac{S_{\triangle C E F}}{S_{\triangle C D F}}=\frac{\frac{3}{4} S_{\triangle A C F}}{\frac{1}{2} S_{\triangle B C F}}=\frac{\frac{3}{4} \times \frac{2}{3}}{\frac{1}{2} \times \frac{1}{3}}=\frac{\frac{1}{2}}{\frac{1}{6}}=3
$$