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1. [2025青海中考]如图，AB是$\odot O$的直径，$∠ CAB = 40°$，则$∠ ADC$的度数是（
A.$80°$
B.$50°$
C.$40°$
D.$25°$
B
）A.$80°$
B.$50°$
C.$40°$
D.$25°$
答案
$\because AB$是$\odot O$的直径,$\therefore ∠ ACB = 90°,\therefore ∠ CAB + ∠ B = 90°. \because ∠ CAB = 40°,\therefore ∠ B = 50°. \because \overset{\frown}{AC} = \overset{\frown}{AC},\therefore ∠ ADC = ∠ B = 50°.$
2. [2025无锡新城中学段考]如图,AB是$\odot O$的直径,点C,D,E都是$\odot O$上的点,则$∠ACE + ∠BDE =$ (
A.$70°$
B.$80°$
C.$90°$
D.$100°$
C
)A.$70°$
B.$80°$
C.$90°$
D.$100°$
答案
连接$AD. \because AB$是$\odot O$的直径,$\therefore ∠ ADB = 90°$,$\therefore ∠ ADE + ∠ BDE = 90°. \because \overset{\frown}{AE} = \overset{\frown}{AE},\therefore ∠ ADE = ∠ ACE$,$\therefore ∠ ACE + ∠ BDE = 90°.$
3. [2025常州中考]如图，AB是$\odot O$的直径，CD是$\odot O$的弦. 若$∠ DCB = 45°$，$AD = 1$，则AB的长度为
$\sqrt{2}$
. 答案
$\because AB$是$\odot O$的直径,$\therefore ∠ ADB = 90°. \because \overset{\frown}{BD} = \overset{\frown}{BD}$,$\therefore ∠ DAB = ∠ DCB = 45°,\therefore ∠ ABD = 90° - ∠ DAB = 45°$,$\therefore BD = AD = 1. \because ∠ ADB = 90°,\therefore AB^2 = AD^2 + BD^2 = 1^2 + 1^2 = 2$,$\therefore AB = \sqrt{2}.$
4. 如图,以△ABC的一边AB为直径的半圆O与AC,BC分别交于点D,E,且$\overset{\frown}{DE} = \overset{\frown}{BE}$. 试判断△ABC的形状,并说明理由.
答案
$△ ABC$是等腰三角形. 理由如下：
连接$AE. \because \overset{\frown}{DE} = \overset{\frown}{BE},\therefore ∠ DAE = ∠ BAE.$
$\because AB$为$\odot O$的直径,$\therefore ∠ AEB = 90°$,
$\therefore ∠ AEC = 90° = ∠ AEB.$
$\because AE = AE,\therefore △ ABE ≌ △ ACE$,
$\therefore AB = AC,\therefore △ ABC$为等腰三角形.
连接$AE. \because \overset{\frown}{DE} = \overset{\frown}{BE},\therefore ∠ DAE = ∠ BAE.$
$\because AB$为$\odot O$的直径,$\therefore ∠ AEB = 90°$,
$\therefore ∠ AEC = 90° = ∠ AEB.$
$\because AE = AE,\therefore △ ABE ≌ △ ACE$,
$\therefore AB = AC,\therefore △ ABC$为等腰三角形.
5.如图，AB是$\odot O$的直径，$∠ ACD = 30°$，过D作$DE ⊥ AB$，垂足为E，DE的延长线交$\odot O$于点F，AB=8，求$∠ DAB$的度数和DF的长.
答案
连接$BD.$
$\because ∠ ACD = 30°,\therefore ∠ B = ∠ ACD = 30°.$
$\because AB$是$\odot O$的直径,$\therefore ∠ ADB = 90°$,
$\therefore ∠ DAB = 90° - ∠ B = 60°.$
$\because ∠ ADB = 90°,∠ B = 30°,AB = 8$,
$\therefore AD = \frac{1}{2}AB = 4.$
$\because ∠ DAB = 60°,DE ⊥ AB$,且$AB$是直径,
$\therefore ∠ ADE = 30°,DE = EF$,
$\therefore AE = \frac{1}{2}AD = 2,DE = \sqrt{AD^2 - AE^2} = 2\sqrt{3}$,
$\therefore DF = 2DE = 4\sqrt{3}.$
$\because ∠ ACD = 30°,\therefore ∠ B = ∠ ACD = 30°.$
$\because AB$是$\odot O$的直径,$\therefore ∠ ADB = 90°$,
$\therefore ∠ DAB = 90° - ∠ B = 60°.$
$\because ∠ ADB = 90°,∠ B = 30°,AB = 8$,
$\therefore AD = \frac{1}{2}AB = 4.$
$\because ∠ DAB = 60°,DE ⊥ AB$,且$AB$是直径,
$\therefore ∠ ADE = 30°,DE = EF$,
$\therefore AE = \frac{1}{2}AD = 2,DE = \sqrt{AD^2 - AE^2} = 2\sqrt{3}$,
$\therefore DF = 2DE = 4\sqrt{3}.$
6. 新情境 [2026 淮安清江浦区月考] 有一个未知圆心的圆形工件需要画出圆心. 暂时只有一块足够大的直角三角板(无刻度)可以使用,为此同学们需要先得到两条不同的直径,下列寻找直径AB的方法正确的是 (
B
)答案
B
变式 [2025 徐州丰县期中] 如图，把三角尺的直角顶点O放在破损玻璃镜的圆周上，两直角边与圆弧分别交于点M，N，量得OM=8 cm，ON=6 cm，则该圆玻璃镜的半径是
5
cm.答案
连接$MN$. 由题意,得$∠ O = 90°$,$\therefore MN$是圆玻璃镜的直径. 在$\mathrm{Rt}△ OMN$中,$OM = 8\ \mathrm{cm}$,$ON = 6\ \mathrm{cm}$,$\therefore MN = \sqrt{6^2 + 8^2} = 10(\mathrm{cm})$,$\therefore$ 该圆玻璃镜的半径是$5\ \mathrm{cm}.$
7. [2025 盐城东台期中] 如图，A,B,E,C 四点都在圆上，AD 是△ABC 的高，∠CAD = ∠EAB，AE 是圆的直径吗？为什么？
答案
$AE$是圆的直径. 理由如下：
连接$BE$,则$∠ E = ∠ C.$
$\because AD$是$△ ABC$的高,
$\therefore ∠ ADC = 90°,\therefore ∠ CAD + ∠ C = 90°.$
$\because ∠ CAD = ∠ EAB,\therefore ∠ EAB + ∠ E = 90°$,
$\therefore ∠ ABE = 90°,\therefore AE$是圆的直径.
连接$BE$,则$∠ E = ∠ C.$
$\because AD$是$△ ABC$的高,
$\therefore ∠ ADC = 90°,\therefore ∠ CAD + ∠ C = 90°.$
$\because ∠ CAD = ∠ EAB,\therefore ∠ EAB + ∠ E = 90°$,
$\therefore ∠ ABE = 90°,\therefore AE$是圆的直径.
8. 四边形ABCD内接于$\odot O$,则$∠ A:∠ B:∠ C:∠ D$可以是 （
A.$1:2:3:4$
B.$1:3:2:4$
C.$1:4:2:3$
D.$1:2:4:3$
D
）A.$1:2:3:4$
B.$1:3:2:4$
C.$1:4:2:3$
D.$1:2:4:3$
答案
$\because$ 四边形$ABCD$内接于$\odot O$,$\therefore ∠ A + ∠ C = 180° = ∠ B + ∠ D$,结合选项知选D.
9. [2026南京玄武区一模] 如图，四边形ABCD是$\odot O$的内接四边形，BC是$\odot O$的直径，$∠ ACB = 18°$，D为$\overset{\frown}{AC}$的中点，则$∠ DAC$的度数是
(
A.$36°$
B.$44°$
C.$52°$
D.$55°$
(
A
)A.$36°$
B.$44°$
C.$52°$
D.$55°$
答案
$\because BC$是$\odot O$的直径,$\therefore ∠ BAC = 90°$,$\therefore ∠ B = 90° - 18° = 72°. \because$ 四边形$ABCD$是$\odot O$的内接四边形,$\therefore ∠ B + ∠ D = 180°$,$\therefore ∠ D = 180° - ∠ B = 180° - 72° = 108°. \because D$为$\overset{\frown}{AC}$的中点,$\therefore \overset{\frown}{AD} = \overset{\frown}{CD}$,$\therefore AD = CD$,$\therefore ∠ DAC = ∠ DCA$,$\therefore ∠ DAC = \frac{180° - 108°}{2} = 36°.$
