(1) 如图 1, 求证: $\angle O D C=\angle O E C$;
(2) 如图 2, 延长 $C E$ 交 $B H$ 于点 $F$, 若 $C D \perp O A$, 求证: $F C=F H$;
(3) 如图 3, 在 (2) 的条件下, 点 $G$ 是 $B H$ 上一点, 连接 $A G, B G, H G, O F$, 若 $A G: B G=5: 3$, $H G=2$, 求 $O F$ 的长.
【答案】 【小问 1 详解】

\begin{aligned} & \therefore O D=\frac{1}{2} O A, O E=\frac{1}{2} O B \\ & \because O A=O B, \\ & \therefore O D=O E \\ & \because \angle B O C=2 \angle C H B, \angle A O C=2 \angle C H B \\ & \therefore \angle A O C=\angle B O C \\ & \because O C=O C \\ & \therefore \triangle \mathrm{COD} \cong \triangle \mathrm{COE}, \\ & \therefore \angle C D O=\angle C E O ; \end{aligned}
【小问 2 详解】

$$\therefore \angle C D O=90^{\circ}$$

\begin{aligned} & \therefore \sin \angle O C E=\frac{O E}{O C}=\frac{1}{2} \\ & \therefore \angle O C E=30^{\circ}, \\ & \therefore \angle C O E=90^{\circ}-\angle O C E=60^{\circ} \end{aligned}

\begin{aligned} & \because \angle H=\frac{1}{2} \angle B O C=\frac{1}{2} \times 60^{\circ}=30^{\circ} \\ & \therefore \angle H=\angle E C O, \\ & \therefore F C=F H \end{aligned}
【小问 3 详解】

\begin{aligned} & \therefore O F \perp C H \\ & \therefore \angle F O H=90^{\circ} \end{aligned}

\begin{aligned} & \therefore \angle A O H=\angle B O H=120^{\circ}, \\ & \therefore A H=B H, \angle A G H=60^{\circ} \\ & \because A G: B G=5: 3 \\ & \text { 设 } A G=5 x, \\ & \therefore B G=3 x \end{aligned}

\begin{aligned} & \because \angle H A M=\angle H B G, \\ & \therefore \triangle H A M \cong \triangle H B G \\ & \therefore M H=G H, \\ & \therefore \triangle M H G \text { 为等边三角形 } \\ & \therefore M G=H G=2 \\ & \because A G=A M+M G, \\ & \therefore 5 x=3 x+2 \\ & \therefore x=1, \\ & \therefore A G=5 \\ & \therefore B G=A M=3, \end{aligned}

\begin{aligned} & M N=\frac{1}{2} G M=\frac{1}{2} \times 2=1, H N=H G \cdot \sin 60^{\circ}=\sqrt{3} \\ & \therefore A N=M N+A M=4, \\ & \therefore H B=H A=\sqrt{N A^2+H N^2}=\sqrt{19} \\ & \because \angle F O H=90^{\circ}, \angle O H F=30^{\circ}, \\ & \therefore \angle O F H=60^{\circ} \\ & \because O B=O H, \\ & \therefore \angle B H O=\angle O B H=30^{\circ}, \\ & \therefore \angle F O B=\angle O B F=30^{\circ} \\ & \therefore O F=B F, \\ & \text { 在 Rt } \triangle O F H \text { 中, } \angle O H F=30^{\circ}, \\ & \therefore H F=2 O F \\ & \therefore H B=B F+H F=3 O F=\sqrt{19}, \\ & \therefore O F=\frac{\sqrt{19}}{3} . \end{aligned}