7. [2026苏州工业园区期中] 如图，四边形ABCD是$\odot O$的内接四边形，AB是$\odot O$的直径，点E在$\overset{\frown}{BAD}$上，$∠ E = 35°$，则$∠ ADC$的度数是 ()

A.$110°$
B.$115°$
C.$120°$
D.$125°$

7 D 连接AC.
∵ $\overset{\frown}{BC} = \overset{\frown}{BC}$,
∴ ∠BAC = ∠E = 35°.
∵ AB是$\odot O$的直径,
∴ ∠ACB = 90°,
∴ ∠CBA = 90° - 35° = 55°.
∵ 四边形ABCD是$\odot O$的内接四边形,
∴ ∠ADC = 180° - 55° = 125°.

8. ▶一题多解 [2025镇江丹阳期中] 如图，△ABC内接于⊙O，∠BAC=30°，BC=√3，则⊙O的半径等于.

8 $\sqrt{3}$ 通解 如图1,作$\odot O$的直径CD,连接BD,则∠CBD = 90°.
∵ ∠D = ∠BAC = 30°,$BC = \sqrt{3}$,
∴ CD = 2BC = $2\sqrt{3}$,
∴ $\odot O$的半径为$\sqrt{3}$.
另解 如图2,连接OB,OC,则∠BOC = 2∠A = 2 × 30° = 60°.
∵ OB = OC,
∴ △OBC是等边三角形,
∴ OC = BC = $\sqrt{3}$,即$\odot O$的半径为$\sqrt{3}$.(本题若∠A是45°或60°时,可过点O作BC的垂线,构造含45°,60°的直角三角形,进而解题)

9. [2025宿迁沭阳月考] 如图，AB是$\odot O$的一条弦，将劣弧沿弦AB翻折，连接AO并延长交翻折后的弧于点C，连接BC.若$AB=6\sqrt{2},BC=3$，则AC的长为.

9 7 如图,延长AC交$\odot O$于点D,过点B作$BH ⊥ AD$于点H,连接BD.
∵ $\overset{\frown}{BC}$和$\overset{\frown}{BD}$是等圆中圆周角∠A所对的弧,
∴ $\overset{\frown}{BC} = \overset{\frown}{BD}$,
∴ BD = BC = 3.
∵ AD是直径,
∴ ∠ABD = 90°,
∴ $AD = \sqrt{AB^2 + BD^2} = 9$.
∵ $\frac{1}{2}AB · BD = \frac{1}{2}AD · BH$,
∴ $BH = \frac{AB · BD}{AD} = 2\sqrt{2}$,
∴ $DH = \sqrt{BD^2 - BH^2} = 1$.
∵ BC = BD,$BH ⊥ AD$,
∴ CH = DH = 1,
∴ AC = AD - CH - DH = 7.

10. [2025安徽中考] 如图, 四边形ABCD的顶点都在半圆O上,AB是半圆O的直径,连接OC,∠DAB + 2∠ABC = 180°.
(1)求证:OC//AD.
(2)若AD=2,BC=2√3,求AB的长.

10 (1)证明:
∵ $\overset{\frown}{AC} = \overset{\frown}{AC}$,
∴ ∠AOC = 2∠ABC.
∵ ∠DAB + 2∠ABC = 180°,
∴ ∠DAB + ∠AOC = 180°,
∴ OC//AD.
(2)解:如图,连接BD,交OC于点E.
∵ AB是半圆O的直径,
∴ ∠ADB = 90°.
∵ OC//AD,
∴ ∠OEB = ∠ADB = 90°,
∴ OC ⊥ BD,
∴ DE = BE.
又
∵ OA = OB,
∴ $OE = \frac{1}{2}AD = \frac{1}{2} × 2 = 1$.
设半圆O的半径为r,则CE = r - 1.
在Rt△OEB中,由勾股定理得$BE^2 = OB^2 - OE^2 = r^2 - 1$.
在Rt△CEB中,由勾股定理得$BE^2 = BC^2 - CE^2 = (2\sqrt{3})^2 - (r - 1)^2$.
∴ $r^2 - 1 = (2\sqrt{3})^2 - (r - 1)^2$,
解得$r_1 = 3$,$r_2 = -2$(舍去),
∴ AB = 2r = 6.

11. [2026南通市启秀中学月考]如图,在$\odot O$中,半径OA,OB互相垂直,点C在劣弧AB上.若$∠ ABC=19°$,则$∠ BAC=$ ()

A.$23°$
B.$24°$
C.$25°$
D.$26°$

11 D 连接OC.
∵ $\overset{\frown}{AC} = \overset{\frown}{AC}$,
∴ ∠ABC = $\frac{1}{2}$∠AOC.
∵ ∠ABC = 19°,
∴ ∠AOC = 38°.
∵ OA ⊥ OB,
∴ ∠AOB = 90°,
∴ ∠BOC = ∠AOB - ∠AOC = 90° - 38° = 52°.
∵ $\overset{\frown}{BC} = \overset{\frown}{BC}$,
∴ ∠BAC = $\frac{1}{2}$∠BOC = 26°.

12. 新情境 用量角器按如图方式测量$∠ ABC$的度数,让$∠ ABC$的顶点恰好在量角器圆弧上,两边分别经过圆弧上的$A,C$两点.若点$A,C$对应的刻度分别为$55°,135°$,则$∠ ABC$的度数为$\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ }$.

12 140° 如图,将图形抽象出来,连接OA,OC,DA,DC,设$\odot O$的直径为EF,由题意可知,∠AOE = 55°,∠EOC = 135°,
∴ ∠AOC = ∠EOC - ∠AOE = 135° - 55° = 80°,
∴ ∠ADC = $\frac{1}{2}$∠AOC = 40°.
∵ 四边形ABCD是$\odot O$的内接四边形,
∴ ∠ABC + ∠ADC = 180°,
∴ ∠ABC = 140°.

13. 一题多解 如图,在$\odot O$中,弦$AC ⊥ BD$于点$E$,连接$AB,BC,CD,OA,OB,OC,OD$.
(1)求证:$∠ AOB + ∠ COD = 180°$.
(2)若$AB = 8$,$CD = 6$,求$\odot O$的直径.

13 (1)证明: 通解 由圆周角定理,得∠AOB = 2∠ACB,∠COD = 2∠CBD.
∵ AC ⊥ BD,
∴ ∠BEC = 90°,
∴ ∠ACB + ∠CBD = 90°,
∴ ∠AOB + ∠COD = 2∠ACB + 2∠CBD = 180°.
另解 如图,延长BO交$\odot O$于点F,连接DF,AD.
∵ BF是$\odot O$的直径,
∴ ∠BDF = 90°,
∴ DF ⊥ BD.
∵ AC ⊥ BD,
∴ AC//DF,
∴ ∠CAD = ∠ADF,
∴ $\overset{\frown}{CD} = \overset{\frown}{AF}$,
∴ ∠COD = ∠AOF.
∵ ∠AOB + ∠AOF = 180°,
∴ ∠AOB + ∠COD = 180°.
(2)解:如图,同(1)中另解作辅助线,并连接AF.
由(1)中另解知$\overset{\frown}{AF} = \overset{\frown}{CD}$,
∴ AF = CD = 6.
∵ BF是$\odot O$的直径,
∴ ∠BAF = 90°,
∴ $BF = \sqrt{AB^2 + AF^2} = \sqrt{8^2 + 6^2} = 10$,
∴ $\odot O$的直径为10.