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1. (2026·福州月考)如图，$△ ABC$内接于$\odot O$，$∠ OBC=40°$，则$∠ A$的度数为 (
A.$40°$
B.$50°$
C.$60°$
D.$80°$
B
)A.$40°$
B.$50°$
C.$60°$
D.$80°$
答案
1. B 解析:
∵$OB=OC, ∠ OBC=40°,\therefore ∠ OBC=∠ OCB=40°,$$\therefore ∠ BOC=100°,\therefore ∠ A=\frac{1}{2}∠ BOC=50°.$故选 B.
∵$OB=OC, ∠ OBC=40°,\therefore ∠ OBC=∠ OCB=40°,$$\therefore ∠ BOC=100°,\therefore ∠ A=\frac{1}{2}∠ BOC=50°.$故选 B.
2. (2026·江门期末)如图,AB是$\odot O$的直径,$∠CAB=40°$,则$∠ADC$的度数是 (
A.$60°$
B.$50°$
C.$40°$
D.$30°$
B
)A.$60°$
B.$50°$
C.$40°$
D.$30°$
答案
2. B 解析:
∵ AB 是$\odot O$的直径, $\therefore ∠ ACB=90°.\because ∠ CAB= 40°,\therefore ∠ ABC=90°-40°=50°.\because ∠ ADC=∠ ABC,\therefore ∠ ADC= 50°.$故选 B.
∵ AB 是$\odot O$的直径, $\therefore ∠ ACB=90°.\because ∠ CAB= 40°,\therefore ∠ ABC=90°-40°=50°.\because ∠ ADC=∠ ABC,\therefore ∠ ADC= 50°.$故选 B.
3. (2026·温州期中)如图,AB是$\odot O$的直径,$∠ ACD=∠ CAB$,$AD=4$,$AC=8$,则$\odot O$的半径为 (
A.2
B.$2\sqrt{5}$
C.4
D.$4\sqrt{5}$
B
)A.2
B.$2\sqrt{5}$
C.4
D.$4\sqrt{5}$
答案
3. B 解析:
∵ $∠ ACD=∠ CAB,\therefore \overset{\frown}{AD}=\overset{\frown}{BC},\therefore AD=BC=4.\because AB$ 是$\odot O$ 的直径, $\therefore ∠ ACB= 90°.$ 在 $Rt △ ACB$ 中, $AB = \sqrt{BC^2+AC^2} = \sqrt{4^2+8^2} = 4\sqrt{5},\therefore \odot O$ 的半径为 $2\sqrt{5}.$故选 B.
归纳总结 两条平行弦所夹的弧相等:AB, CD 为弦且$CD// AB,$则$\overset{\frown}{AC}=\overset{\frown}{BD}.$如图,可分情况证明.
4. (2026·盘锦期中)如图,一圆形玻璃镜面被损坏了一部分,为了得到同样大小的镜面,工人师傅用直角尺量得$AB=4\ \mathrm{dm},BC=3\ \mathrm{dm}$,则该圆形镜面的直径为
5
dm.答案
4. 5 解析:如图,连接 AC,
∵ $∠ ABC=90°,$且$∠ ABC$ 是圆周角,
∴ AC 是圆形镜面的直径.$\because AB=4\ \mathrm{dm},BC=3\ \mathrm{dm},\therefore AC= \sqrt{AB^2+BC^2} = \sqrt{4^2+3^2} = 5(\mathrm{dm}).$
5. (2026·苏州期中)如图,OA是$\odot O$的半径,以OA为直径的$\odot C$与$\odot O$的弦AB相交于点D,则AD
=
BD(填“＞”“=”或“＜”).答案
5. = 解析:如图,连接 OD,
∵ OA 为$\odot C$的直径,
∴ $∠ ADO= 90°,\therefore OD ⊥ AB,\therefore AD=BD.$
6. (2026·赤峰期末)如图,AB为$\odot O$的直径,点C在$\odot O$上,且$CO ⊥ AB$于点O,弦CD与AB相交于点E,连接AD,若$∠ A=25°$,则$∠ AEC$的度数为
70°
.答案
6. 70° 解析:
∵ AB 为$\odot O$的直径,点 C 在$\odot O$上,且 $CO ⊥ AB$ 于点 O, $\therefore \overset{\frown}{AC}=\overset{\frown}{BC},\therefore ∠ ADC = \frac{1}{2}∠ AOC.$
∵ $CO ⊥ AB$, $\therefore ∠ AOC=90°,\therefore ∠ ADC = \frac{1}{2}∠ AOC = 45°,\therefore ∠ AEC = ∠ A+ ∠ ADC=25°+45°=70°.$
∵ AB 为$\odot O$的直径,点 C 在$\odot O$上,且 $CO ⊥ AB$ 于点 O, $\therefore \overset{\frown}{AC}=\overset{\frown}{BC},\therefore ∠ ADC = \frac{1}{2}∠ AOC.$
∵ $CO ⊥ AB$, $\therefore ∠ AOC=90°,\therefore ∠ ADC = \frac{1}{2}∠ AOC = 45°,\therefore ∠ AEC = ∠ A+ ∠ ADC=25°+45°=70°.$
7. 如图，AE是$\odot O$的直径，半径$OC ⊥$弦$AB$，点$D$为垂足，连接$BE$，$BC$.
(1)若$∠ BEC=25°$，求$∠ AOC$的度数；
(2)若$∠ CEA=∠ A$，$EC=6$，求$\odot O$的半径.
(1)若$∠ BEC=25°$，求$∠ AOC$的度数；
(2)若$∠ CEA=∠ A$，$EC=6$，求$\odot O$的半径.
答案
7. (1)
∵ $OC ⊥ AB,\therefore \overset{\frown}{AC}=\overset{\frown}{BC},\therefore ∠ AEC = ∠ CEB = 25°,$ $\therefore ∠ AOC=2∠ AEC=50°.$
(2)连接 AC.
∵ AE 是$\odot O$的直径,
∴ $∠ ABE = ∠ ACE = 90°$, $\therefore ∠ AEB+∠ BAE = 90°.\because ∠ CEA = ∠ BAE, ∠ CEB = ∠ AEC,$ $\therefore ∠ BAE = ∠ AEC = ∠ CEB = 30°,\therefore AC = \frac{1}{2} AE.$ 又
∵ $AE^2 = AC^2+EC^2,\therefore AE=4\sqrt{3},\therefore \odot O$ 的半径为 $2\sqrt{3}.$
归纳总结 (1)有直径造直角或有直角作直径,是利用直径解题的常用方法.
(2)解决圆中线段计算问题的常用方法是构造含有此线段的特殊三角形,如直角三角形或等腰三角形.
∵ $OC ⊥ AB,\therefore \overset{\frown}{AC}=\overset{\frown}{BC},\therefore ∠ AEC = ∠ CEB = 25°,$ $\therefore ∠ AOC=2∠ AEC=50°.$
(2)连接 AC.
∵ AE 是$\odot O$的直径,
∴ $∠ ABE = ∠ ACE = 90°$, $\therefore ∠ AEB+∠ BAE = 90°.\because ∠ CEA = ∠ BAE, ∠ CEB = ∠ AEC,$ $\therefore ∠ BAE = ∠ AEC = ∠ CEB = 30°,\therefore AC = \frac{1}{2} AE.$ 又
∵ $AE^2 = AC^2+EC^2,\therefore AE=4\sqrt{3},\therefore \odot O$ 的半径为 $2\sqrt{3}.$
归纳总结 (1)有直径造直角或有直角作直径,是利用直径解题的常用方法.
(2)解决圆中线段计算问题的常用方法是构造含有此线段的特殊三角形,如直角三角形或等腰三角形.
8.（福建中考）如图，四边形ABCD内接于$\odot O$，$AB=CD$，A为$\overset{\frown}{BD}$的中点，$∠ BDC = 60°$，则$∠ ADB$等于（
A.$40°$
B.$50°$
C.$60°$
D.$70°$
A
）A.$40°$
B.$50°$
C.$60°$
D.$70°$
答案
8. A 解析:连接 OA,OB,OD,OC.
∵ $∠ BDC=60°,\therefore ∠ BOC= 2∠ BDC=120°.\because AB=DC,\therefore ∠ AOB=∠ DOC.\because A$ 为$\overset{\frown}{BD}$的中点,$\therefore \overset{\frown}{AB}=\overset{\frown}{AD},\therefore ∠ AOB=∠ AOD,\therefore ∠ AOB=∠ AOD=∠ DOC= \frac{1}{3}×(360°-∠ BOC)=80°,\therefore ∠ ADB=\frac{1}{2}∠ AOB=40°.$
∵ $∠ BDC=60°,\therefore ∠ BOC= 2∠ BDC=120°.\because AB=DC,\therefore ∠ AOB=∠ DOC.\because A$ 为$\overset{\frown}{BD}$的中点,$\therefore \overset{\frown}{AB}=\overset{\frown}{AD},\therefore ∠ AOB=∠ AOD,\therefore ∠ AOB=∠ AOD=∠ DOC= \frac{1}{3}×(360°-∠ BOC)=80°,\therefore ∠ ADB=\frac{1}{2}∠ AOB=40°.$
