10. 教材练习变式 [2026 扬州梅岭中学期末] 如图,点 A,B,C在$\odot O$上,点 B 在$\overset{\frown}{AC}$上,连接 OA,OC,CB,作射线 AB,已知$∠AOC=116°$,则$∠CBD$的度数是 ()

A.$58°$
B.$116°$
C.$64°$
D.$60°$

如图,在优弧$\overset{\frown}{AC}$上取一点$E$,连接$AE$,$CE$. $\because \overset{\frown}{AC} = \overset{\frown}{AC}$,$\therefore ∠ E = \frac{1}{2}∠ AOC = \frac{1}{2} × 116° = 58°. \because$ 四边形$ABCE$是$\odot O$的内接四边形,$\therefore ∠ E + ∠ ABC = 180°$. 又$\because ∠ CBD + ∠ ABC = 180°$,$\therefore ∠ CBD = ∠ E = 58°.$

11. [2025扬州仪征二模]如图，点A,B,C,D,E在$\odot O$上.若$∠ E+∠ C=145°$,则$\overset{\frown}{AB}$的度数为（）

A.$25°$
B.$35°$
C.$50°$
D.$70°$

连接$BC. \because$ 四边形$BCDE$是$\odot O$的内接四边形,$\therefore ∠ E + ∠ BCD = 180°$,即$∠ E + ∠ ACD + ∠ ACB = 180°. \because ∠ E + ∠ ACD = 145°$,$\therefore ∠ ACB = 35°$,$\therefore \overset{\frown}{AB}$的度数为$70°.$

12. [2026宿迁宿豫区期中]如图,四边形ABCD内接于$\odot O$,若四边形OABC是菱形,则$∠D=$°。

$\because$ 四边形$ABCD$内接于$\odot O$,$\therefore ∠ B + ∠ D = 180°.$
$\because$ 四边形$OABC$是菱形,$\therefore ∠ B = ∠ AOC$,$\therefore ∠ AOC + ∠ D = 180°$. 又$\because ∠ AOC = 2∠ D$,$\therefore ∠ D = 60°.$

13. [2025无锡江阴长泾二中月考] 如图,四边形ABCD内接于$\odot O$,$\overset{\frown}{AB} = \overset{\frown}{AC}$.若$∠ BAC = 50°$,求$∠ D$的度数.

$\because \overset{\frown}{AB} = \overset{\frown}{AC}$,$\therefore ∠ ACB = ∠ B.$
$\because ∠ BAC + ∠ B + ∠ ACB = 180°$,$∠ BAC = 50°$,
$\therefore ∠ B = ∠ ACB = \frac{1}{2}(180° - ∠ BAC) = 65°.$
$\because$ 四边形$ABCD$内接于$\odot O$,
$\therefore ∠ B + ∠ D = 180°$,$\therefore ∠ D = 180° - ∠ B = 115°.$

14. [2025宿迁经开区月考] 如图,以$△ ABC$的一边AB为直径的半圆与其他两边AC,BC分别交于点D,E,且E为$\overset{\frown}{BD}$的中点.若AB=16, BC=8,则BD的长为 ()

A.7
B.7.5
C.$2\sqrt{14}$
D.$2\sqrt{15}$

如图,连接$AE. \because AB$是直径,$\therefore ∠ ADB = ∠ AEB = 90°$,$\therefore AE ⊥ BC$,$BD ⊥ AC. \because E$为$\overset{\frown}{BD}$的中点,$\therefore \overset{\frown}{DE} = \overset{\frown}{EB}$,$\therefore ∠ CAE = ∠ EAB. \because ∠ C + ∠ CAE = 90°$,$∠ ABE + ∠ EAB = 90°$,$\therefore ∠ C = ∠ ABC$,$\therefore AC = AB = 16$. 又$\because AE ⊥ BC$,$\therefore BE = CE = 4$,$\therefore AE = \sqrt{AB^2 - BE^2} = 4\sqrt{15}. \because \frac{1}{2}AC · BD = \frac{1}{2}BC · AE$,$\therefore BD = \frac{BC · AE}{AC} = \frac{8 × 4\sqrt{15}}{16} = 2\sqrt{15}.$

15. [2026宿迁泗洪期中]如图，已知四边形ABCD内接于$\odot O$，连接AC，记$∠ BAC$的度数为$α$，$∠ CAD$的度数为$β$. 若$AB = AC$，$AB // CD$，则有（）

A.$2α + 3β = 180°$
B.$3α + 4β = 360°$
C.$3α + 2β = 180°$
D.$4α + 3β = 360°$

$\because AB = AC$,$\therefore ∠ B = ∠ ACB. \because ∠ B + ∠ ACB + ∠ BAC = 180°$,$\therefore 2∠ ACB + α = 180°$,$\therefore ∠ ACB = 90° - \frac{1}{2}α. \because AB // CD$,$\therefore ∠ ACD = ∠ BAC = α$,$\therefore ∠ BCD = ∠ ACB + ∠ ACD = 90° - \frac{1}{2}α + α = 90° + \frac{1}{2}α. \because$ 四边形$ABCD$内接于$\odot O$,$\therefore ∠ BAD + ∠ BCD = 180°$. 又$\because ∠ BAD = ∠ BAC + ∠ CAD = α + β$,$\therefore 90° + \frac{1}{2}α + α + β = 180°$,整理,得$3α + 2β = 180°.$

16. [2026 泰州泰兴月考] 已知AB是$\odot O$的直径，$\overset{\frown}{AC} = \overset{\frown}{CD}$，点D不与点A，B重合，直线AD与直线BC相交于点E，$∠ BAE = 26°$，则$∠ CAE$的度数为。

设$∠ CAE = α$. 如图1,当$\overset{\frown}{AD}$是劣弧时,$\because \overset{\frown}{AC} = \overset{\frown}{CD}$,$\therefore ∠ CAD = ∠ D = ∠ B = α. \because AB$是$\odot O$的直径,$\therefore ∠ ACB = 90°$,$\therefore ∠ CAB + ∠ B = 90°$,$\therefore α + 26° + α = 90°$,解得$α = 32°.$
如图2,当$\overset{\frown}{AD}$是优弧时,同理可证$∠ CAD = ∠ CDA = ∠ ABC = α$,$∠ CAB + ∠ ABC = 90°$,$\therefore ∠ CAD - ∠ DAB + ∠ ABC = 90°$.
$\therefore α - 26° + α = 90°$,解得$α = 58°$. 综上可知,$∠ CAE$的度数为$32°$或$58°.$

17. [2026 连云港赣榆区月考] ，∠ACB=90°，∠ACB的平分线交⊙O于点D。若CD=3，AC=√2，则AB的长为。

如图,过点$A$作$AE ⊥ DC$,垂足为$E$,连接$DB$,$AD$,则$∠ AEC = ∠ AED = 90°. \because CD$平分$∠ ACB$,$\therefore ∠ ACD = ∠ BCD = 45°$,$\therefore △ ACE$是等腰直角三角形.$\because AC = \sqrt{2}$,$\therefore AE = CE = 1$,$\therefore DE = CD - CE = 3 - 1 = 2$. 在$\mathrm{Rt}△ ADE$中,由勾股定理,得$AD = \sqrt{AE^2 + DE^2} = \sqrt{1^2 + 2^2} = \sqrt{5}. \because ∠ ACB = 90°$,$\therefore AB$为$\odot O$的直径,$\therefore ∠ ADB = 90°. \because \overset{\frown}{AD} = \overset{\frown}{AD}$,$\therefore ∠ ABD = ∠ ACD = 45°$,$\therefore △ ADB$是等腰直角三角形,$\therefore AB = \sqrt{AD^2 + BD^2} = \sqrt{2}AD = \sqrt{2} × \sqrt{5} = \sqrt{10}.$

18.如图，A，P，B，C是$\odot O$上四点，$∠ APC = ∠ CPB = 60°$.
(1)判断$△ ABC$的形状，并证明你的结论.
(2)当点P位于什么位置时，四边形PBOA是菱形？并说明理由.
(3)探究三条线段PA，PB，PC之间的数量关系，并说明理由.

(1)$△ ABC$是等边三角形. 证明如下：
$\because \overset{\frown}{BC} = \overset{\frown}{BC}$,$\therefore ∠ BAC = ∠ CPB.$
$\because \overset{\frown}{AC} = \overset{\frown}{AC}$,$\therefore ∠ ABC = ∠ APC.$
$\because ∠ APC = ∠ CPB = 60°$,$\therefore ∠ ABC = ∠ BAC = 60°$,
$\therefore △ ABC$为等边三角形.
(2)当点$P$位于$\overset{\frown}{AB}$的中点时,四边形$PBOA$是菱形. 理由如下：
连接$OP.$
$\because P$是$\overset{\frown}{AB}$的中点,$\therefore \overset{\frown}{AP} = \overset{\frown}{BP}$,
$\therefore ∠ AOP = ∠ BOP.$
$\because ∠ AOB = 2∠ ACB = 120°$,$\therefore ∠ AOP = ∠ BOP = 60°.$
又$\because OA = OP = OB$,$\therefore △ OAP$,$△ OBP$都是等边三角形,
$\therefore OA = AP = OB = PB$,
$\therefore$ 四边形$PBOA$是菱形.
(3)$PA + PB = PC$. 理由如下：
如图,在$PC$上截取$PD = AP.$
$\because ∠ APC = 60°$,$\therefore △ APD$是等边三角形,
$\therefore AD = AP = PD$,$∠ ADP = 60°$,$\therefore ∠ ADC = 120°.$
$\because ∠ APB = ∠ APC + ∠ BPC = 120°$,
$\therefore ∠ ADC = ∠ APB.$
在$△ APB$和$△ ADC$中,$\begin{cases} ∠ APB = ∠ ADC, \\ ∠ ABP = ∠ ACD, \\ AP = AD, \end{cases}$
$\therefore △ APB ≌ △ ADC(\mathrm{AAS})$,$\therefore BP = CD.$
又$\because AP = PD$,$\therefore AP + BP = CD + PD$,
即$PA + PB = PC.$