9. 如图，AC, BC 为$\odot O$的两条弦，D, G 分别为AC, BC 的中点，$\odot O$的半径为 2. 若$∠ ACB = 45°$, 则 DG 的长为 ()

A.2
B.$\sqrt{3}$
C.$\frac{3}{2}$
D.$\sqrt{2}$

9 D 如图,连接 $AO,BO,AB. \because ∠ ACB = 45°,\therefore ∠ AOB = 2∠ ACB = 90°. \because \odot O$ 的半径为 2,$\therefore AO = BO = 2,\therefore AB = 2\sqrt{2}.$$\because D,G$ 分别是 AC,BC 的中点,$\therefore DG$ 是 $△ ABC$ 的中位线,$\therefore DG = \frac{1}{2}AB = \sqrt{2}.$

10. [2026镇江丹徒区月考]如图，$∠ C = 15°$，且$\overset{\frown}{AB} = \overset{\frown}{BC} = \overset{\frown}{CD}$，则$∠ E$的度数为（）

A.$30°$
B.$35°$
C.$40°$
D.$45°$

10 C $\because ∠ C = 15°,\therefore \overset{\frown}{AD}$ 的度数为 $30°,\therefore \overset{\frown}{ABD}$ 的度数为 $330°. \because \overset{\frown}{AB} = \overset{\frown}{BC} = \overset{\frown}{CD},\therefore \overset{\frown}{BC}$ 的度数为 $110°,\therefore ∠ BAC = 55°.$$\because ∠ BAC$ 是 $△ ACE$ 的外角,$\therefore ∠ BAC = ∠ C + ∠ E,\therefore ∠ E = ∠ BAC - ∠ C = 55° - 15° = 40°.$

11. [2026南京理工大学附中月考]如图，AB是$\odot O$的弦，$∠ OAB = 20°$，则弦AB所对的圆周角的度数为。

11 70°或110° $\because AO = BO,\therefore ∠ OBA = ∠ OAB = 20°,$$\therefore ∠ AOB = 180° - ∠ OBA - ∠ OAB = 180° - 20° - 20° = 140°,$$\therefore$ 弦 AB 将 $\odot O$ 分成的两段弧的度数分别为 $140°,220°,$$\therefore$ 弦 AB 所对圆周角的度数为 $70°$ 或 $110°.$

12. [2025苏州吴中区一模]如图,点A,B,C,D,E在$\odot O$上,D是$\overset{\frown}{AB}$的中点,$CB = CE$.若$∠ AOB = 100°$,$∠ OBC = 55°$,则$∠ DCE$的度数为.

12 85° 连接 OC. $\because ∠ AOB = 100°,\therefore \overset{\frown}{AB}$ 的度数为 $100°.$$\because D$ 是 $\overset{\frown}{AB}$ 的中点,$\therefore \overset{\frown}{BD}$ 的度数为 $50°. \because OB = OC,\therefore ∠ OCB = ∠ OBC = 55°,$$\therefore ∠ BOC = 180° - ∠ OCB - ∠ OBC = 180° - 55° - 55° = 70°,$$\therefore \overset{\frown}{BC}$ 的度数为 $70°. \because CB = CE,\therefore \overset{\frown}{BC} = \overset{\frown}{CE},\therefore \overset{\frown}{CE}$ 的度数为 $70°,$$\therefore \overset{\frown}{DE}$ 的度数为 $360° - 50° - 70° × 2 = 170°,$$\therefore ∠ DCE$ 的度数为 $85°.$

13.［2025扬州邗江区期中］如图，C，D是半圆ACB上的两点，点P是直径AB上一点，若∠APC=∠BPD，则称∠CPD是$\overset{\frown}{CD}$的“幸运角”.如图，若直径AB=2，弦CE⊥AB，$\overset{\frown}{CD}$的“幸运角”∠CPD为90°，则CD的长为.

13 $\sqrt{2}$ 如图,连接 $OC,OD. \because \overset{\frown}{CD}$ 的“幸运角”$∠ CPD$ 为 $90°,$$\therefore ∠ APC = ∠ BPD = \frac{1}{2}(180° - 90°) = 45°,\therefore ∠ APE = ∠ BPD = 45°. \because CE ⊥ AB,$$\therefore ∠ E = 90° - ∠ APE = 90° - 45° = 45°,$$\therefore ∠ COD = 2∠ E = 90°. \because$ 直径 $AB = 2,\therefore CO = DO = 1.$ 在 $\mathrm{Rt}△ OCD$ 中,由勾股定理得 $CD = \sqrt{OC^2 + OD^2} = \sqrt{1^2 + 1^2} = \sqrt{2}.$

14. [2025南京竹山中学月考]在$\odot O$中,设$\overset{\frown}{BC}$所对的圆周角为$∠ BAC$. 求证: $∠ BAC = \frac{1}{2}∠ BOC$.
证明:圆心$O$可能在$∠ BAC$的一边上、内部或外部(如图1,图2和图3).
如图1,当圆心$O$在$∠ BAC$的一边上时,
$\because OA = OC, \therefore ∠ A = ∠ C$.
$\because ∠ BOC = ∠ A + ∠ C$,
$\therefore ∠ BOC = 2∠ A$,即$∠ BAC = \frac{1}{2}∠ BOC$.
请你完成图2、图3的证明.

14 证明:如图1,连接 AO 并延长交 $\odot O$ 于点 D.
由题图1的证明可知$∠ BAO = \frac{1}{2}∠ BOD,∠ CAO = \frac{1}{2}∠ COD,$
$\therefore ∠ BAC = ∠ BAO + ∠ CAO = \frac{1}{2}∠ BOD + \frac{1}{2}∠ COD = \frac{1}{2}(∠ BOD + ∠ COD) = \frac{1}{2}∠ BOC.$
如图2,延长 BO 交 $\odot O$ 于点 E,连接 CE,
$\therefore ∠ E = ∠ BAC.$
由题图1的证明可知$∠ E = \frac{1}{2}∠ BOC,$
$\therefore ∠ BAC = \frac{1}{2}∠ BOC.$

15. 推理能力 在四边形ABCD中,用无刻度的直尺和圆规完成下列作图(保留作图痕迹,不写作法,写出必要的文字说明).
(1)如图1,连接BD,在CD边上作点M,使得$∠ AMB = ∠ ADB$;
(2)如图2,在CD边上作点N,使得$∠ BND = ∠ A$.

15 解:(1)如图1,作AB和AD的垂直平分线,它们相交于点O,然后以点O为圆心,OA长为半径作圆交CD于点M,则点M为所作.
(2)如图2,作点A关于BD的对称点A',再作DA'和BD的垂直平分线,它们相交于点O,然后以点O为圆心,OB长为半径作圆交CD于点N,则点N为所作.