(1) 若 $\odot O$ 的半径为 5 , 求 $C G$ 的长;
(2) 试探究 $D E$ 与 $E F$ 之间的数量关系, 写出并证明你的结论. (请用两种证法解答)
【答案】 解: (1) 如答案图 2 . 连接 $C E$.
\begin{aligned} & \because \overparen{C E}=\overparen{C E}, \therefore \angle C O E=2 \angle C G E . \\ & \because \angle D O E=2 \angle C G E, \therefore \angle C O E=\angle D O E . \\ & \because A B \text { 为 } \odot O \text { 的切线, } C \text { 为切点, } \therefore O C \perp A B \\ & \therefore \angle O C B=90^{\circ} \therefore D F \perp A B, \text { 垂足为 } F, \end{aligned}

\begin{aligned} & \therefore \angle D F B=90^{\circ}, \therefore \angle O C B=\angle D F B=90^{\circ}, \therefore O C / / D F, \\ & \therefore \angle C O E=\angle O E D, \therefore \angle D O E=\angle O E D, \therefore O D=D E . \because O D=O E, \\ & \therefore \triangle O D E \text { 是等边三角形, } \therefore \angle D O E=60^{\circ}, \therefore \angle C G E=30^{\circ} . \\ & \because \odot O \text { 的半径为 } 5, \therefore G E=10 . \because G E \text { 是 } \odot O \text { 的直径, } \therefore \angle G C E=90^{\circ}, \text { 。 } \\ & \therefore \text { 在Rt } \triangle G C E \text { 中, } G C=G E \cdot \cos \angle C G E=10 \times \cos 30^{\circ}=5 \sqrt{3} . \end{aligned}
(2) $D E=2 E F$.

$\because O C=O E, \therefore \triangle O C E$ 为等边三角形, $\therefore \angle O C E=60^{\circ}$.
$\because \angle O C B=90^{\circ}, \therefore \angle E C F=30^{\circ}$.
$\therefore$ 在 Rt $\triangle C E F$ 中, $E F=\frac{1}{2} C E, \therefore E F=\frac{1}{2} D E$, 即 $D E=2 E F,$

$\therefore \angle O H F=90^{\circ} \therefore \angle O C B=\angle D F C=90^{\circ}, \therefore$ 四边形 $O C F H$ 是矩形,

\begin{aligned} & \therefore C F=O H . \because \triangle O D E \text { 是等边三角形, } \therefore D E=O E . \\ & \because O H \perp D F, \therefore D H=E H . \\ & \because \angle C O E=\angle D O E, \therefore \overparen{C E}=\overparen{D E}, \therefore C E=D E, \\ & \therefore C E=O E . \because C F=O H, \therefore \mathrm{Rt} \triangle C F E \cong \mathrm{Rt} \triangle O H E, \\ & \therefore E F=E H, \quad \therefore D H=E H=E F, \therefore D E=2 E F \end{aligned}