9. [2025 北京海淀月考,★☆]如图,已知 AB 是$\odot O$的直径,半径$OC⊥AB$,D 是 CO 的中点,若过点 D 的弦$EF// AB$,则下列结论正确的是 ( )
A.$\overset{\frown }{EC}= \overset{\frown }{AE}$
B.$\overset{\frown }{EC}= 2\overset{\frown }{AE}$
C.$\overset{\frown }{EF}= 3\overset{\frown }{AE}$
D.$\overset{\frown }{ABF}= 4\overset{\frown }{AE}$
A.$\overset{\frown }{EC}= \overset{\frown }{AE}$
B.$\overset{\frown }{EC}= 2\overset{\frown }{AE}$
C.$\overset{\frown }{EF}= 3\overset{\frown }{AE}$
D.$\overset{\frown }{ABF}= 4\overset{\frown }{AE}$
答案：
B 如图，连接 OE、OF、CE，$\because OC⊥AB$，$EF// AB$，$\therefore OC⊥EF$，$\therefore ∠ODE = 90^{\circ}$，又$\because D$是 CO 的中点，$\therefore CE = OE = OC$，$\therefore \triangle COE$是等边三角形，$\therefore ∠EOD = 60^{\circ}$，$\because EF// AB$，$\therefore ∠AOE = ∠OED = 90^{\circ} - ∠EOD = 30^{\circ}$，$\therefore \overset{\frown}{EC} = 2\overset{\frown}{AE}$，故选项 A 错误，选项 B 正确；易得$∠EOF = 120^{\circ}$，$\therefore \overset{\frown}{EF} = 4\overset{\frown}{AE}$，故选项 C 错误；$\because \overset{\frown}{ABF}$所对的圆心角的度数为$210^{\circ}$，$\therefore \overset{\frown}{ABF} = 7\overset{\frown}{AE}$，故选项 D 错误.故选 B.
B 如图，连接 OE、OF、CE，$\because OC⊥AB$，$EF// AB$，$\therefore OC⊥EF$，$\therefore ∠ODE = 90^{\circ}$，又$\because D$是 CO 的中点，$\therefore CE = OE = OC$，$\therefore \triangle COE$是等边三角形，$\therefore ∠EOD = 60^{\circ}$，$\because EF// AB$，$\therefore ∠AOE = ∠OED = 90^{\circ} - ∠EOD = 30^{\circ}$，$\therefore \overset{\frown}{EC} = 2\overset{\frown}{AE}$，故选项 A 错误，选项 B 正确；易得$∠EOF = 120^{\circ}$，$\therefore \overset{\frown}{EF} = 4\overset{\frown}{AE}$，故选项 C 错误；$\because \overset{\frown}{ABF}$所对的圆心角的度数为$210^{\circ}$，$\therefore \overset{\frown}{ABF} = 7\overset{\frown}{AE}$，故选项 D 错误.故选 B.
10. [学科特色 方程思想 ★☆]如图,AB 为$\odot O$的直径,点 D 是$\overset{\frown }{AC}$的中点,过点 D 作$DE⊥AB$于点 E,延长 DE 交$\odot O$于点 F.若$AC= 4\sqrt {3}$,$AE= 2$,则$\odot O$的直径为 (
A.$8\sqrt {3}$
B.8
C.10
D.$8\sqrt {2}$
B
)A.$8\sqrt {3}$
B.8
C.10
D.$8\sqrt {2}$
答案：B 连接 OF（图略），$\because DE⊥AB$，$\therefore DE = EF$，$\overset{\frown}{AD} = \overset{\frown}{AF}$，$\because$点 D 是$\overset{\frown}{AC}$的中点，$\therefore \overset{\frown}{AD} = \overset{\frown}{CD}$，$\therefore \overset{\frown}{AC} = \overset{\frown}{DF}$，$\therefore DF = AC = 4\sqrt{3}$，$\therefore EF = \frac{1}{2}DF = 2\sqrt{3}$，设$OA = OF = x$，$\therefore OE = x - 2$，在$Rt\triangle OEF$中，$x^{2} = (2\sqrt{3})^{2} + (x - 2)^{2}$，解得$x = 4$，$\therefore AB = 2x = 8$.故选 B.
11. [★☆]如图,点 A、B、C、D 在$\odot O$上,且$\overset{\frown }{AD}= \overset{\frown }{BC}$,E 是 AB 延长线上一点,且$BE= AB$,F 是 EC 的中点,若$BF= 6cm$,则$BD= $______
12
cm.答案：答案 12
解析 连接 AC（图略），$\because F$是 EC 的中点，$\therefore CF = EF$，$\because BE = AB$，$\therefore BF$是$\triangle EAC$的中位线，$\therefore BF = \frac{1}{2}AC$，$\because \overset{\frown}{AD} = \overset{\frown}{BC}$，$\therefore \overset{\frown}{AD} + \overset{\frown}{AB} = \overset{\frown}{BC} + \overset{\frown}{AB}$，$\therefore \overset{\frown}{DB} = \overset{\frown}{AC}$，$\therefore BD = AC$，$\therefore BF = \frac{1}{2}BD$，$\therefore BD = 2BF = 12cm$.
解析 连接 AC（图略），$\because F$是 EC 的中点，$\therefore CF = EF$，$\because BE = AB$，$\therefore BF$是$\triangle EAC$的中位线，$\therefore BF = \frac{1}{2}AC$，$\because \overset{\frown}{AD} = \overset{\frown}{BC}$，$\therefore \overset{\frown}{AD} + \overset{\frown}{AB} = \overset{\frown}{BC} + \overset{\frown}{AB}$，$\therefore \overset{\frown}{DB} = \overset{\frown}{AC}$，$\therefore BD = AC$，$\therefore BF = \frac{1}{2}BD$，$\therefore BD = 2BF = 12cm$.
12. [2025 江苏泰州姜堰期中,★☆]如图,在$\odot O$中,$\overset{\frown }{AB}= 2\overset{\frown }{BC}$,且$BD⊥OC$,垂足为 D.若$AB= 8$,$CD= 2$,则$\odot O$的半径为______.
答案：
答案 5
解析 如图，过点 O 作 AB 的垂线交 AB 于点 E，交$\overset{\frown}{AB}$于点 F，连接 OB.$\because OF⊥AB$，$AB = 8$，$\therefore \overset{\frown}{AF} = \overset{\frown}{BF} = \frac{1}{2}\overset{\frown}{AB}$，$AE = BE = \frac{1}{2}AB = \frac{1}{2}×8 = 4$，$\because \overset{\frown}{AB} = 2\overset{\frown}{BC}$，$\therefore \overset{\frown}{BC} = \frac{1}{2}\overset{\frown}{AB} = \overset{\frown}{BF}$，$\therefore ∠BOC = ∠BOF$，即 OB 平分$∠COF$，$\because BD⊥OC$，$\therefore BD = BE = 4$，设$\odot O$的半径为 r，则$OB = OC = r$，$\because CD = 2$，$\therefore OD = OC - CD = r - 2$，在$Rt\triangle BOD$中，由勾股定理，得$BD^{2} + OD^{2} = OB^{2}$，$\therefore 4^{2} + (r - 2)^{2} = r^{2}$，$\therefore r = 5$，$\therefore \odot O$的半径为 5.
答案 5
解析 如图，过点 O 作 AB 的垂线交 AB 于点 E，交$\overset{\frown}{AB}$于点 F，连接 OB.$\because OF⊥AB$，$AB = 8$，$\therefore \overset{\frown}{AF} = \overset{\frown}{BF} = \frac{1}{2}\overset{\frown}{AB}$，$AE = BE = \frac{1}{2}AB = \frac{1}{2}×8 = 4$，$\because \overset{\frown}{AB} = 2\overset{\frown}{BC}$，$\therefore \overset{\frown}{BC} = \frac{1}{2}\overset{\frown}{AB} = \overset{\frown}{BF}$，$\therefore ∠BOC = ∠BOF$，即 OB 平分$∠COF$，$\because BD⊥OC$，$\therefore BD = BE = 4$，设$\odot O$的半径为 r，则$OB = OC = r$，$\because CD = 2$，$\therefore OD = OC - CD = r - 2$，在$Rt\triangle BOD$中，由勾股定理，得$BD^{2} + OD^{2} = OB^{2}$，$\therefore 4^{2} + (r - 2)^{2} = r^{2}$，$\therefore r = 5$，$\therefore \odot O$的半径为 5.
13. [2025 贵州贵阳期中,★☆]如图,AB 为$\odot O$的弦,半径 OC,OD 分别交 AB 于点 E,F,且$\overset{\frown }{AC}= \overset{\frown }{DB}$.
(1)求证：$AE= BF$.
(2)作半径$ON⊥AB$于点 M,若$AB= 12$,$MN= 3$,求 OM 的长.
(1)求证：$AE= BF$.
(2)作半径$ON⊥AB$于点 M,若$AB= 12$,$MN= 3$,求 OM 的长.
答案：
解析 (1)证明：连接 OA，OB，如图.$\because OA = OB$，$\therefore ∠OAB = ∠OBA$，$\because \overset{\frown}{AC} = \overset{\frown}{BD}$，$\therefore ∠AOE = ∠BOF$，在$\triangle AOE$和$\triangle BOF$中，$\begin{cases}∠OAE = ∠OBF,\\OA = OB,\\∠AOE = ∠BOF,\end{cases}$ $\therefore \triangle AOE ≌ \triangle BOF(ASA)$，$\therefore AE = BF$.(2)连接 OA，如图.$\because OM⊥AB$，$\therefore AM = \frac{1}{2}AB = 6$，设$OM = x$，则$OA = ON = x + 3$，在$Rt\triangle AOM$中，由勾股定理得$6^{2} + x^{2} = (x + 3)^{2}$，解得$x = 4.5$，$\therefore OM = 4.5$.
解析 (1)证明：连接 OA，OB，如图.$\because OA = OB$，$\therefore ∠OAB = ∠OBA$，$\because \overset{\frown}{AC} = \overset{\frown}{BD}$，$\therefore ∠AOE = ∠BOF$，在$\triangle AOE$和$\triangle BOF$中，$\begin{cases}∠OAE = ∠OBF,\\OA = OB,\\∠AOE = ∠BOF,\end{cases}$ $\therefore \triangle AOE ≌ \triangle BOF(ASA)$，$\therefore AE = BF$.(2)连接 OA，如图.$\because OM⊥AB$，$\therefore AM = \frac{1}{2}AB = 6$，设$OM = x$，则$OA = ON = x + 3$，在$Rt\triangle AOM$中，由勾股定理得$6^{2} + x^{2} = (x + 3)^{2}$，解得$x = 4.5$，$\therefore OM = 4.5$.
14. [新课标 推理能力]如图,点 C、D 三等分以 O 为圆心的$\overset{\frown }{MN}$,连接 MN、CD、OC、OD、OM,下列结论错误的是 ( )
A.$∠COM= ∠COD$
B.若$OM= MN$,则$∠COD= 20^{\circ }$
C.$MN// CD$
D.$MN= 3CD$
A.$∠COM= ∠COD$
B.若$OM= MN$,则$∠COD= 20^{\circ }$
C.$MN// CD$
D.$MN= 3CD$
答案：
D 如图，连接 ON、MC、DN，过点 O 作$OE⊥CD$，交$\overset{\frown}{CD}$于点 E，$\because \overset{\frown}{CM} = \overset{\frown}{CD}$，$\therefore ∠COM = ∠COD$，故选项 A 结论正确.当$OM = MN$时，$\because OM = ON$，$\therefore OM = ON = MN$，$\therefore \triangle OMN$为等边三角形，$\therefore ∠MON = 60^{\circ}$，$\because \overset{\frown}{CM} = \overset{\frown}{CD} = \overset{\frown}{DN}$，$\therefore ∠COM = ∠COD = ∠DON$，$\therefore ∠COD = 20^{\circ}$，故选项 B 结论正确.$\because OE⊥CD$，$\therefore \overset{\frown}{CE} = \overset{\frown}{DE}$，$\therefore \overset{\frown}{ME} = \overset{\frown}{NE}$，$\therefore OE⊥MN$，$\therefore MN// CD$，故选项 C 结论正确.$\because \overset{\frown}{CM} = \overset{\frown}{CD} = \overset{\frown}{DN}$，$\therefore MC = CD = DN$，$\because MC + CD + DN ＞ MN$，$\therefore MN ＜ 3CD$，故选项 D 结论错误.故选 D.
D 如图，连接 ON、MC、DN，过点 O 作$OE⊥CD$，交$\overset{\frown}{CD}$于点 E，$\because \overset{\frown}{CM} = \overset{\frown}{CD}$，$\therefore ∠COM = ∠COD$，故选项 A 结论正确.当$OM = MN$时，$\because OM = ON$，$\therefore OM = ON = MN$，$\therefore \triangle OMN$为等边三角形，$\therefore ∠MON = 60^{\circ}$，$\because \overset{\frown}{CM} = \overset{\frown}{CD} = \overset{\frown}{DN}$，$\therefore ∠COM = ∠COD = ∠DON$，$\therefore ∠COD = 20^{\circ}$，故选项 B 结论正确.$\because OE⊥CD$，$\therefore \overset{\frown}{CE} = \overset{\frown}{DE}$，$\therefore \overset{\frown}{ME} = \overset{\frown}{NE}$，$\therefore OE⊥MN$，$\therefore MN// CD$，故选项 C 结论正确.$\because \overset{\frown}{CM} = \overset{\frown}{CD} = \overset{\frown}{DN}$，$\therefore MC = CD = DN$，$\because MC + CD + DN ＞ MN$，$\therefore MN ＜ 3CD$，故选项 D 结论错误.故选 D.
15. [新课标 几何直观 2025 河南驻马店西平期中]如图,CD 是$\odot O$的直径,点 A 是半圆上靠近点 D 的一个三等分点,B 是$\overset{\frown }{AD}$的中点,P 为直线 CD 上的一个动点,当$CD= 4$时,$AP+BP$的最小值为______.
答案：
答案 $2\sqrt{2}$
解析 如图，作点 A 关于 CD 的对称点$A'$，连接$A'B$，交 CD 于点 P，此时$PA + PB$的值最小，最小值为$A'B$的长，连接$OA'$，OB，$\because$点 A 与$A'$关于 CD 对称，点 A 是半圆上靠近点 D 的一个三等分点，$\therefore PA = PA'$，$∠A'OD = ∠AOD = 60^{\circ}$，$\because$点 B 是$\overset{\frown}{AD}$的中点，$\therefore ∠BOD = 30^{\circ}$，$\therefore ∠A'OB = ∠A'OD + ∠BOD = 90^{\circ}$，易知$OB = OA' = 2$，$\therefore A'B = 2\sqrt{2}$，$\therefore PA + PB = PA' + PB = A'B = 2\sqrt{2}$，$\therefore AP + BP$的最小值为$2\sqrt{2}$.
答案 $2\sqrt{2}$
解析 如图，作点 A 关于 CD 的对称点$A'$，连接$A'B$，交 CD 于点 P，此时$PA + PB$的值最小，最小值为$A'B$的长，连接$OA'$，OB，$\because$点 A 与$A'$关于 CD 对称，点 A 是半圆上靠近点 D 的一个三等分点，$\therefore PA = PA'$，$∠A'OD = ∠AOD = 60^{\circ}$，$\because$点 B 是$\overset{\frown}{AD}$的中点，$\therefore ∠BOD = 30^{\circ}$，$\therefore ∠A'OB = ∠A'OD + ∠BOD = 90^{\circ}$，易知$OB = OA' = 2$，$\therefore A'B = 2\sqrt{2}$，$\therefore PA + PB = PA' + PB = A'B = 2\sqrt{2}$，$\therefore AP + BP$的最小值为$2\sqrt{2}$.