1. In $\triangle A B C$, point $D$ is on $B C$, $\frac{B D}{D C}=\frac{1}{3}, E, G$ are on $A B, A D$ respectively, $\frac{A E}{E B}=\frac{2}{3}, \frac{A G}{G D}=\frac{1}{2}, E G$ intersects $A C$ at point $F$, find $\frac{A F}{F C}$.
1. Extend $C B, F E$ to intersect at $H$. For $\triangle A D B$ and the transversal $G E H$, we have $\frac{A G}{G D} \cdot \frac{D H}{H B} \cdot \frac{B E}{E A}=\frac{1}{2} \cdot \frac{D H}{H B} \cdot \frac{3}{2}=1$, so $\frac{D H}{H B}=\frac{4}{3}$, which means $\frac{C H}{H D}=\frac{7}{4}$. For $\triangle A C D$ and the transversal $F G H$, we have $\frac{A F}{F C} \cdot \frac{C H}{H D} \cdot \frac{D G}{G A}=\frac{A F}{F C} \cdot \frac{7}{4} \cdot \frac{2}{1}=1$, solving for $\frac{A F}{F C}=\frac{2}{7}$.