13. $A B C D$ is a trapezium with $A B / / D C$ and $A B>D C$. $E$ is a point on $A B$ such that $A E=D C$. AC meets $D E$ and $D B$ at $F$ and $G$ respectively. Find the value of $\frac{A B}{C D}$ for which $\frac{\text { Area of } \triangle D F G}{\text { Area of trapezium } A B C D}$ is maximum.
(2 marks)
$A B C D$ 是梯形, 其中 $A B / / D C$, 且 $A B>D C \circ E$ 是 $A B$ 上的一點, 使得 $A E=$ $D C \circ A C$ 分別交 $D E$ 和 $D B$ 於 $F$ 和 $G \circ$ 求 $\frac{A B}{C D}$, 使得 $\frac{\triangle D F G \text { 的面積 }}{\text { 形 } A B C D \text { 的面積 }}$ 達至最大値。
(2 分)
13. 3
13. We denote by $[X Y Z]$ the area of $X Y Z$. By scaling we may assume $A E=D C=1$ and $E B=x$. Since $\triangle C D G$ $\sim \triangle A B G$, we have $\frac{C G}{A G}=\frac{1}{1+x}$. Note that $A E C D$ is a parallelogram, so $A F=F C$ and hence
$$
\frac{F G}{A C}=\frac{\frac{1}{2}[1+(1+x)]-1}{1+(1+x)}=\frac{x}{4+2 x} .
$$

From this we get
$$
\frac{[D F G]}{[A B C D]}=\frac{\frac{F G}{A C} \cdot[A C D]}{2[A C E]+x[A C E]}=\frac{x}{(4+2 x)(2+x)}=\frac{1}{2\left(\frac{2}{\sqrt{x}}+\sqrt{x}\right)^{2}}=\frac{1}{2\left(\frac{2}{\sqrt{x}}-\sqrt{x}\right)^{2}+8}
$$

This is maximum when $\frac{2}{\sqrt{x}}=\sqrt{x}$, or $x=2$, in which case $\frac{A B}{C D}=1+x=3$.