14. Let $A B C D$ be a convex quadrilateral and $E$ be a point on $C D$ such that the circumcircle of $\triangle A B E$ is tangent to $C D$. Suppose $A C$ meets $B E$ at $F, B D$ meets $A E$ at $G$, and $A C$ meets $B D$ at $H$. If $F G / / C D$, and the areas of $\triangle A B H, \triangle B C E$ and $\triangle A D E$ are 2,3 and 4 respectively, find the area of $\triangle A B E$.
(2 marks)
設 $A B C D$ 為凸四邊形, $E$ 為 $C D$ 上的一點, 使得 $\triangle A B E$ 的外接圓與 $C D$ 相切。設 $A C$ 與 $B E$ 交於 $F ; B D$ 與 $A E$ 交於 $G ; A C$ 與 $B D$ 交於 $H \circ$ 若 $F G / /$ $C D$ ，且 $\triangle A B H 、 \triangle B C E$ 和 $\triangle A D E$ 的面積分別是 2、3、4, 求 $\triangle A B E$ 的面積。（2 分）
14. $1+\sqrt{15}$
14. Let $[P Q R]$ denote the area of $\triangle P Q R$ and $[A B E]=x$. Then $\frac{B F}{F E}=\frac{B G}{G D}=\frac{[A B E]}{[A D E]}=\frac{x}{4}$ and $\frac{A G}{G E}=\frac{A F}{F C}=\frac{[A B E]}{[C B E]}=\frac{x}{3}$.
This yields $[A F E]=\frac{F E}{B E}[A B E]=\frac{4 x}{x+4}$ and similarly $[B G E]=\frac{G E}{A E}[A B E]=\frac{3 x}{x+3}$.
Using Menelaus' Theorem, we have $\frac{A H}{H F} \cdot \frac{F B}{B E} \cdot \frac{E G}{G A}=1$, i.e. $\frac{A H}{H F} \cdot \frac{x}{x+4} \cdot \frac{3}{x}=1$ or $\frac{A H}{H F}=\frac{x+4}{3}$. This gives
$$
[A H E]=\frac{A H}{A F}[A F E]=\frac{4 x}{x+7} .
$$

Similarly, $\frac{B H}{H G} \cdot \frac{G A}{A E} \cdot \frac{E F}{F B}=1$ implies $\frac{B H}{H G} \cdot \frac{x}{x+3} \cdot \frac{4}{x}=1$ or $\frac{B H}{H G}=\frac{x+3}{4}$. This gives
$$
[B H E]=\frac{B H}{B G}[B G E]=\frac{3 x}{x+7} .
$$

Using the relation $[A B H]+[A H E]+[B H E]=[A B E]$, we have $2+\frac{4 x}{x+7}+\frac{3 x}{x+7}=x$. This simplifies to the quadratic equation $x^{2}-2 x-14=0$, which has a unique positive solution $1+\sqrt{15}$.

Remark. The condition that $C D$ is tangent to the circumcircle of $\triangle A B E$ is not needed. In fact it can be shown that subject to the remaining given conditions such tangency must hold.