2. $A B C$ is an acute-angled triangle. $D$ is a point on $B C$ such that $B D: D C=2: 3$, while $E$ is a point on $A C$ such that $A E: E C=3: 4 . A D$ and $B E$ meet at $F$. Find $\frac{A F}{F D} \times \frac{B F}{F E}$.
(1 mark) $A B C$ 是銳角三角形。 $D$ 是 $B C$ 上的一點, 使得 $B D: D C=2: 3 \circ E$ 是 $A C$ 上的一點,使得 $A E: E C=3: 4 \circ A D$ 交 $B E$ 於 $F$ 。求 $\frac{A F}{F D} \times \frac{B F}{F E}$ 。
2. $\frac{35}{12}$
2. Produce $A D$ to $G$ and $B E$ to $H$ such that $A C / / B G$ and $B C / / A H$. Let $B D=2 x, D C=3 x, A E=3 y$, $E C=4 y$. Since $\triangle A C D \sim \triangle G B D$, we have $\frac{A C}{B G}=\frac{A D}{D G}=\frac{C D}{D B}=\frac{3}{2} \quad$. Hence $\quad B G=\frac{14 y}{3} \quad$ and $A D: D G=3: 2$, i.e. $A D=\frac{3}{5} A G$. On the other hand, since $\triangle A E F \sim \triangle G B F, \frac{A F}{F G}=\frac{F E}{B F}=\frac{A E}{B G}=\frac{3 y}{\frac{14 y}{3}}=\frac{9}{14}$. Thus $\frac{B F}{F E}=\frac{14}{9}$ and $A F=\frac{9}{23} F G$. If follows that $F D=A D-A F=\frac{3}{5} A G-\frac{9}{23} F G=\frac{24}{115} F G$. Thus $\frac{A F}{F D}=\frac{\frac{9}{23} F G}{\frac{24}{115} F G}=\frac{15}{8}$ and hence $\frac{A F}{F D} \times \frac{B F}{F E}=\frac{15}{8} \times \frac{14}{9}=\frac{35}{12}$.