1. [2026 扬州梅岭中学教育集团月考] 如图,在$\odot O$中, $AB=CD$,则下列结论错误的是 ()

A.$\overset{\frown}{AB}=\overset{\frown}{CD}$
B.$\overset{\frown}{AC}=\overset{\frown}{BD}$
C.$AC=BD$
D.$AD=BD$

1 D $\because AB = CD,\therefore \overset{\frown}{AB}=\overset{\frown}{CD}$(结论A不符合题意),$\therefore \overset{\frown}{AB}-\overset{\frown}{AD}=\overset{\frown}{CD}-\overset{\frown}{AD}$,即$\overset{\frown}{BD}=\overset{\frown}{AC}$(结论B不符合题意),$\therefore BD = AC$(结论C不符合题意),而$AD = BD$通过已知条件无法得出,故选D.

2. [2026 盐城盐都区月考] 如图，AB，CD 是$\odot O$的直径，$\overset{\frown}{AE}=\overset{\frown}{BD}$。若$∠ AOE=32°$，则$∠ COE$的度数为（）

A.$32°$
B.$48°$
C.$60°$
D.$64°$

2 D $\because \overset{\frown}{AE}=\overset{\frown}{BD},\therefore ∠BOD = ∠AOE = 32°,\therefore ∠AOC = ∠BOD = 32°,\therefore ∠COE = ∠AOE + ∠AOC = 64°.$

3. [2025苏州工业园区期中]如图，OA,OB,OC均为$\odot O$的半径，$∠ BOD=22°$，$AO ⊥ BO$，点D为$\overset{\frown}{AC}$的中点，则$∠ BOC$的度数是°。

3 46 $\because AO ⊥ BO,\therefore ∠AOB = 90°.\because ∠BOD = 22°,\therefore ∠AOD = 68°.\because$ 点 D 为$\overset{\frown}{AC}$的中点,$\therefore \overset{\frown}{AD}=\overset{\frown}{CD},\therefore ∠COD = ∠AOD = 68°,\therefore ∠BOC = ∠COD - ∠BOD = 68° - 22° = 46°.$

4. 给出下列说法:①弧长相等的弧所对的弦相等;②相等的圆心角所对的弧相等;③在同圆或等圆中,同弧或等弧所对的圆心角相等;④相等的弦所对的圆心角相等.其中说法正确的是.(填序号)

4 ③ 在同圆或等圆中,如果两个圆心角、两条弧、两条弦中有一组量相等,那么它们所对应的其余各组量都分别相等,所以③正确;①②④没有说明是在同圆或等圆中,故错误.

5.［2025南京玄武区期末］如图，AB是$\odot O$的直径，点C，D在$\odot O$上，$OC// BD$.求证：$AC=CD$.

5 证明:如图,连接 OD.
$\because OC// BD,\therefore ∠1 = ∠3,∠2 = ∠B.$
$\because OD = OB,\therefore ∠B = ∠3,$
$\therefore ∠1 = ∠2,\therefore AC = CD.$

6. [2026 扬州邗江中学月考] 如图，AB是$\odot O$的直径，$\overset{\frown}{BC} = \overset{\frown}{CD} = \overset{\frown}{DE}$，$∠ BOC = 40°$，则$\overset{\frown}{AE}$的度数为。

6 60° $\because \overset{\frown}{BC}=\overset{\frown}{CD}=\overset{\frown}{DE},\therefore ∠BOC = ∠COD = ∠DOE = 40°.$
$\because AB$ 为直径,$\therefore ∠AOE = 180° - 3×40° = 60°,\therefore \overset{\frown}{AE}$的度数为60°.

7. [2026南京建邺区期中]如图,在$△ ABC$中,以B为圆心,BA为半径画$\overset{\frown}{AE}$分别交AC,BC于点D,E.若$CD=AB$,$∠ B=87°$,则$\overset{\frown}{DE}$的度数为.

7 31° 如图,连接 BD.$\because AB = BD,\therefore ∠DAB = ∠ADB.\because CD = AB,AB = BD,\therefore CD = BD,\therefore ∠DBC = ∠C.\because ∠ADB = ∠DBC + ∠C = 2∠DBC,\therefore ∠DAB = 2∠DBC.\because ∠ABC = 87°,∠BAC + ∠C + ∠ABC = 180°,\therefore 3∠DBC = 93°,\therefore ∠DBC = 31°,\therefore \overset{\frown}{DE}$的度数为31°.

8. 如图,以平行四边形ABCD的顶点A为圆心,AB长为半径作$\odot A$,分别交BC,AD于点E,F,交BA的延长线于点G,连接EG.
(1)求证:$\overset{\frown}{EF} = \overset{\frown}{FG}$.
(2)若$\overset{\frown}{EG}$的度数为$100°$,求$∠ EGB$的度数.

8 (1)证明:如图,连接 AE.
$\because$ 四边形 ABCD 是平行四边形,
$\therefore AD// BC,$
$\therefore ∠EAF = ∠AEB,∠GAF = ∠GBE.$
$\because AE = AB,$
$\therefore ∠GBE = ∠AEB,\therefore ∠EAF = ∠GAF,\therefore \overset{\frown}{EF}=\overset{\frown}{FG}.$
(2)解:$\because \overset{\frown}{EG}$的度数为$100°$,
$\therefore ∠EAG = 100°,\therefore ∠BAE = 80°.$
$\because AE = AG,\therefore ∠AEG = ∠AGE,$
$\because ∠BAE = ∠AEG + ∠AGE,$
$\therefore ∠EGB = \frac{1}{2}∠BAE = 40°.$