17. As shown in the figure, $\triangle \mathrm{AED}$ is a right-angled triangle, with the lengths of the two legs being $\left\{\begin{array}{l}A E=15 \\ D E=20\end{array}\right.$ . A square $\mathrm{ABCD}$ is constructed with $\mathrm{AD}$ as one side, and a parallelogram $\mathrm{ABFE}$ is constructed with $\mathrm{AB}$ and $\mathrm{AE}$ as sides. $\mathrm{EF}$ intersects $\mathrm{AD}$ at point $\mathrm{G}$, and $\mathrm{AC}$ intersects $\mathrm{FG}$ at point $\mathrm{H}$. The difference in area (larger area minus smaller area) between $\triangle A G H$ and $\triangle C F H$ is $\qquad$ -
【Analysis】By the Pythagorean theorem, we know that $A D^{2}=15^{2}+20^{2}=625$, so $\mathrm{AD}$ is 25 and $S_{\triangle A D E}=\frac{1}{2} \times A E \times E D=\frac{1}{2} \times A D \times E G$, so $E G=15 \times 20 \div 25=12$. Again by the Pythagorean theorem, $G D^{2}=E D^{2}-E G^{2}=20^{2}-12^{2}=256$, so $\mathrm{GD}$ is 16. Since $E F=A B=A D=25$, then $G F=13$.
$$
\begin{array}{l}
S_{\triangle F H C}+S_{G H C D}=S_{F G C D}=(13+25) \times 16 \div 2=304 \\
S_{\triangle A G H}+S_{G H C D}=S_{\triangle A D C}=25 \times 25 \div 2=312.5 \\
\text { Therefore, } S_{\triangle A G H}-S_{\triangle F H C}=312.5-304=8.5
\end{array}
$$