17. In the diagram below, $A B C D$ is a square. The points $A, B$ and $G$ are collinear. The line segments $A C$ and $D G$ intersect at $E$, and the line segments $D G$ and $B C$ intersect at $F$. Suppose that $D E=15 \mathrm{~cm}, E F=9 \mathrm{~cm}$, and $F G=x \mathrm{~cm}$. Find the value of $x$.
17. Ans: 16
Let $A B=y \mathrm{~cm}$. Since $\triangle D C F \sim \triangle G B F$, we have
$$
\frac{D C}{D F}=\frac{G B}{G F} \Longrightarrow \frac{y}{24}=\frac{G B}{x} \Longrightarrow G B=\frac{x y}{24} .
$$

Since $\triangle D C E \sim \triangle G A E$, we have
$$
\frac{D C}{D E}=\frac{G A}{G E} \Longrightarrow \frac{y}{15}=\frac{y+G B}{9+x} \Longrightarrow G B=\frac{y(9+x)}{15}-y
$$

Hence we have
$$
\frac{x y}{24}=\frac{y(9+x)}{15}-y \Longrightarrow \frac{x}{24}=\frac{9+x}{15}-1 \Longrightarrow x=16
$$