(1)求证: $F D / / A B$;
(2) 若 $A C=2 \sqrt{5}, B C=\sqrt{5}$, 求 $F D$ 的长.
【答案】 24. 解: (1) 连接 $\mathrm{AD} 、 \mathrm{BD} 、 \mathrm{OD}, ~ \because \mathrm{AB} \odot 0$ 的为直径, $\therefore \angle \mathrm{ACB}=\angle \mathrm{ADB}=90^{\circ}$,
$\because \mathrm{CD}$ 平分 $\angle \mathrm{ACB}, \therefore \angle \mathrm{ACD}=45^{\circ}, \therefore \angle \mathrm{ABD}=45^{\circ}, \therefore \triangle \mathrm{ABD}$ 为等腰直角三角形, $\because 0$ 为 $\mathrm{AB}$ 中点, $\therefore \mathrm{OD} \perp \mathrm{AB}, \because \mathrm{DF}$ 为 $\odot 0$ 的切线, $\therefore \mathrm{DF} \perp \mathrm{OD}, \therefore \mathrm{DF} / / \mathrm{AB}$;
(2) 过 $C$ 作 $C G \perp A B$ 于 $G$, 由勾股定理知 $A B=\sqrt{A C^2+B C^2}=\sqrt{20+5}=5, \therefore O D=\frac{5}{2}$,
\begin{aligned} & \because \sin \angle \mathrm{ABC}=\frac{\mathrm{AC}}{\mathrm{AB}}-\frac{2 \sqrt{5}}{5}, \cos \angle \mathrm{ABC}=\frac{\mathrm{BC}}{\mathrm{AB}}-\frac{\sqrt{5}}{5}, \\ & \therefore \sin \angle \mathrm{GBC}=\frac{\mathrm{CG}}{\mathrm{BC}}=\sin \angle \mathrm{ABC}=\frac{2 \sqrt{5}}{5}, \cos \angle \mathrm{GBC}=\frac{\mathrm{BG}}{\mathrm{BC}}=\cos \angle \mathrm{ABC}=\frac{\sqrt{5}}{5} \\ & \therefore \mathrm{CG}=\frac{2 \sqrt{5}}{5} \mathrm{BC}=2, \mathrm{BG}=\frac{\sqrt{5}}{5} \mathrm{BC}=1,0 \mathrm{OG}=\mathrm{OB}-\mathrm{BG}=\frac{5}{2}-1=\frac{3}{2}, \\ & \text { 由 (1) 知 } \mathrm{FD} / / \mathrm{AB}, \therefore \angle \mathrm{COG}=\angle \mathrm{OFD}, \text { 又 } \because \angle \mathrm{ODF}=\angle \mathrm{CGO}, \therefore \triangle \mathrm{ODF} \operatorname{}=\triangle \mathrm{CGO}, \\ & \therefore \frac{\mathrm{CG}}{\mathrm{OD}}=\frac{\mathrm{OG}}{\mathrm{FD}}, \therefore \mathrm{FD}=\frac{\mathrm{OG} \times \mathrm{OD}}{\mathrm{CG}}-\frac{3}{2} \times \frac{5}{2}-\frac{15}{8} . \end{aligned}