1. 教材练习变式 下列图形中的角是圆周角的是（）


A.①②③
B.②④
C.②⑤
D.②

1 D 顶点在圆上,并且两边都和圆相交的角叫作圆周角.观察题图可知只有②符合.

2. [2026泰州靖江期末] 如图，AB是$\odot O$的直径，点C，D在$\odot O$上且位于AB的异侧.若$∠ BCO = 65°$,则$∠ D$的度数为 （）

A.$65°$
B.$35°$
C.$30°$
D.$25°$

2 D $\because OC = OB,\therefore ∠ CBO = ∠ BCO = 65°,\therefore ∠ BOC = 180° -∠ CBO - ∠ BCO = 50°,\therefore ∠ D = \frac{1}{2}∠ BOC = \frac{1}{2} × 50° = 25°.$

3. [2024赤峰中考]如图，AD是$\odot O$的直径，AB是$\odot O$的弦，半径$OC ⊥ AB$，连接CD，交OB于点E，$∠ BOC = 42°$，则$∠ OED$的度数是（）

A.$61°$
B.$63°$
C.$65°$
D.$67°$

3 B $\because$ 半径 $OC ⊥ AB,\therefore \overset{\frown}{AC} = \overset{\frown}{BC},\therefore ∠ AOC = ∠ BOC = 42°,$$\therefore ∠ D = \frac{1}{2}∠ AOC = 21°. \because OC = OD,\therefore ∠ C = ∠ D = 21°,$$\therefore ∠ OED = ∠ C + ∠ BOC = 21° + 42° = 63°.$

4. [2026 徐州沛县模拟] 船在航行过程中，通常通过测定角度来确定是否会遇到暗礁. 如图，点A,B表示两个灯塔，暗礁分布在经过A,B两点的$\odot O$区域内，优弧AB上任一点C都是有触礁危险的临界点，$∠ ACB$就是“危险角”. 已知$∠ AOB = 110°$，要保证船D安全航行，则$∠ D$的度数可能是（）

A.$65°$
B.$60°$
C.$55°$
D.$50°$

4 D 设AD与$\odot O$的另一个公共点为 E,连接 BE. $\because \overset{\frown}{AB} = \overset{\frown}{AB},\therefore ∠ AEB = \frac{1}{2}∠ AOB = \frac{1}{2} × 110° = 55°. \because ∠ AEB$ 是$△ BDE$ 的外角,$\therefore ∠ AEB ＞ ∠ D,\therefore ∠ D ＜ 55°.$ 只有选项 D 符合题意.

5. [2026 盐城东台期中] 如图,在$\odot O$中,$OA ⊥ BC$,$∠ ADB=30°$,$BC=6\sqrt{3}$,则OC的长为.

第5题图 第6题图

5 6 如图,设 AO 与 BC 交于点 E,连接 OB. $\because \overset{\frown}{AB} = \overset{\frown}{AB},$$\therefore ∠ AOB = 2∠ ADB = 2 × 30° = 60°. \because OA ⊥ BC,\therefore \overset{\frown}{AC} = \overset{\frown}{AB},$$CE = \frac{1}{2}BC = \frac{1}{2} × 6\sqrt{3} = 3\sqrt{3},∠ OEC = 90°,\therefore ∠ AOC = ∠ AOB = 60°,\therefore ∠ OCE = 90° - ∠ AOC = 90° - 60° = 30°.$ 在 $\mathrm{Rt}△ OCE$ 中,$\because ∠ OCE = 30°,\therefore OE = \frac{1}{2}OC.$ 设 $OE = x$,则 $OC = 2x.$$\because ∠ OEC = 90°,\therefore OE^2 + CE^2 = OC^2,\therefore x^2 + (3\sqrt{3})^2 = (2x)^2,$解得 $x=3$(负值舍去),$\therefore OC = 6.$

6.如图,某博览会上有一圆形展示区,在其圆形边缘的点P处安装了一台监视器,它的监控角度是$55°$.为了监控整个展区,最少需要在圆形边缘上共安装这样的监视器台.

6 4 $\because ∠ P = 55°,\therefore ∠ P$ 所对弧所对的圆心角是 $110°.$$\because 360° ÷ 110° = 3\frac{3}{11},\therefore$ 最少需要在圆形边缘上共安装这样的监视器 4 台.

7. ▶一题多解 如图,等腰三角形ABC中,AB=AC,⊙O过点B,C,且与AB,AC分别相交于点D,E.求证:BD=CE.

7 证明:
通解 $\because AB = AC,\therefore ∠ B = ∠ C,\therefore \overset{\frown}{BE} = \overset{\frown}{CD}.$$\therefore \overset{\frown}{BE} - \overset{\frown}{DE} = \overset{\frown}{CD} - \overset{\frown}{DE},$即$\overset{\frown}{BD} = \overset{\frown}{CE}.$$\therefore BD = CE.$
另解 连接 $OB,OC,OD,OE.$$\because AB = AC,\therefore ∠ ABC = ∠ ACB.$又$\because ∠ COD = 2∠ ABC,∠ BOE = 2∠ ACB,$$\therefore ∠ BOE = ∠ COD,\therefore \overset{\frown}{BDE} = \overset{\frown}{DEC},$$\therefore \overset{\frown}{BDE} - \overset{\frown}{DE} = \overset{\frown}{DEC} - \overset{\frown}{DE},\therefore \overset{\frown}{BD} = \overset{\frown}{CE},\therefore BD = CE.$

8. [2025无锡惠山区期中] 如图，AB是$\odot O$的直径，弦$CD ⊥ AB$于点E，点P在$\odot O$上，$∠ 1 = ∠ C$。
(1) 求证：$CB // PD$。
(2) 若$BC = 3$，$∠ C = 30°$，求$\odot O$的直径。

8 (1)证明:$\because \overset{\frown}{BD} = \overset{\frown}{BD},\therefore ∠ P = ∠ C,$$\because ∠ 1 = ∠ C,$$\therefore ∠ 1 = ∠ P,\therefore CB // PD.$
(2)解:连接 $OC.$$\because ∠ ECB = 30°,CE ⊥ BE,\therefore ∠ CBE = 60°.$$\because OC = OB,$$\therefore △ BOC$ 为等边三角形,$\therefore OB = BC = 3,\therefore \odot O$ 的直径为 6.