9. [2025 北京海淀月考,★☆]如圖,已知 AB 是$\odot O$的直徑,半徑$OC⊥AB$,D 是 CO 的中點,若過點 D 的弦$EF// AB$,則下列結(jié)論正確的是 ( )
A.$\overset{\frown }{EC}= \overset{\frown }{AE}$
B.$\overset{\frown }{EC}= 2\overset{\frown }{AE}$
C.$\overset{\frown }{EF}= 3\overset{\frown }{AE}$
D.$\overset{\frown }{ABF}= 4\overset{\frown }{AE}$
答案：B 如圖，連接 OE、OF、CE，$\because OC⊥AB$，$EF// AB$，$\therefore OC⊥EF$，$\therefore ∠ODE = 90^{\circ}$，又$\because D$是 CO 的中點，$\therefore CE = OE = OC$，$\therefore \triangle COE$是等邊三角形，$\therefore ∠EOD = 60^{\circ}$，$\because EF// AB$，$\therefore ∠AOE = ∠OED = 90^{\circ} - ∠EOD = 30^{\circ}$，$\therefore \overset{\frown}{EC} = 2\overset{\frown}{AE}$，故選項 A 錯誤，選項 B 正確；易得$∠EOF = 120^{\circ}$，$\therefore \overset{\frown}{EF} = 4\overset{\frown}{AE}$，故選項 C 錯誤；$\because \overset{\frown}{ABF}$所對的圓心角的度數(shù)為$210^{\circ}$，$\therefore \overset{\frown}{ABF} = 7\overset{\frown}{AE}$，故選項 D 錯誤.故選 B.
10. [學(xué)科特色 方程思想 ★☆]如圖,AB 為$\odot O$的直徑,點 D 是$\overset{\frown }{AC}$的中點,過點 D 作$DE⊥AB$于點 E,延長 DE 交$\odot O$于點 F.若$AC= 4\sqrt {3}$,$AE= 2$,則$\odot O$的直徑為 (
B
)
A.$8\sqrt {3}$
B.8
C.10
D.$8\sqrt {2}$
答案：B 連接 OF（圖略），$\because DE⊥AB$，$\therefore DE = EF$，$\overset{\frown}{AD} = \overset{\frown}{AF}$，$\because$點 D 是$\overset{\frown}{AC}$的中點，$\therefore \overset{\frown}{AD} = \overset{\frown}{CD}$，$\therefore \overset{\frown}{AC} = \overset{\frown}{DF}$，$\therefore DF = AC = 4\sqrt{3}$，$\therefore EF = \frac{1}{2}DF = 2\sqrt{3}$，設(shè)$OA = OF = x$，$\therefore OE = x - 2$，在$Rt\triangle OEF$中，$x^{2} = (2\sqrt{3})^{2} + (x - 2)^{2}$，解得$x = 4$，$\therefore AB = 2x = 8$.故選 B.
11. [★☆]如圖,點 A、B、C、D 在$\odot O$上,且$\overset{\frown }{AD}= \overset{\frown }{BC}$,E 是 AB 延長線上一點,且$BE= AB$,F 是 EC 的中點,若$BF= 6cm$,則$BD= $______
12
cm.
答案：答案 12
解析 連接 AC（圖略），$\because F$是 EC 的中點，$\therefore CF = EF$，$\because BE = AB$，$\therefore BF$是$\triangle EAC$的中位線，$\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$.
12. [2025 江蘇泰州姜堰期中,★☆]如圖,在$\odot O$中,$\overset{\frown }{AB}= 2\overset{\frown }{BC}$,且$BD⊥OC$,垂足為 D.若$AB= 8$,$CD= 2$,則$\odot O$的半徑為______.
答案：答案 5
解析 如圖，過點 O 作 AB 的垂線交 AB 于點 E，交$\overset{\frown}{AB}$于點 F，連接 OB.$\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$，設(shè)$\odot O$的半徑為 r，則$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$的半徑為 5.
13. [2025 貴州貴陽期中,★☆]如圖,AB 為$\odot O$的弦,半徑 OC,OD 分別交 AB 于點 E,F,且$\overset{\frown }{AC}= \overset{\frown }{DB}$.
(1)求證：$AE= BF$.
(2)作半徑$ON⊥AB$于點 M,若$AB= 12$,$MN= 3$,求 OM 的長.
答案：解析 (1)證明：連接 OA，OB，如圖.$\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)連接 OA，如圖.$\because OM⊥AB$，$\therefore AM = \frac{1}{2}AB = 6$，設(shè)$OM = x$，則$OA = ON = x + 3$，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. [新課標 推理能力]如圖,點 C、D 三等分以 O 為圓心的$\overset{\frown }{MN}$,連接 MN、CD、OC、OD、OM,下列結(jié)論錯誤的是 ( )
A.$∠COM= ∠COD$
B.若$OM= MN$,則$∠COD= 20^{\circ }$
C.$MN// CD$
D.$MN= 3CD$
答案：D 如圖，連接 ON、MC、DN，過點 O 作$OE⊥CD$，交$\overset{\frown}{CD}$于點 E，$\because \overset{\frown}{CM} = \overset{\frown}{CD}$，$\therefore ∠COM = ∠COD$，故選項 A 結(jié)論正確.當$OM = MN$時，$\because OM = ON$，$\therefore OM = ON = MN$，$\therefore \triangle OMN$為等邊三角形，$\therefore ∠MON = 60^{\circ}$，$\because \overset{\frown}{CM} = \overset{\frown}{CD} = \overset{\frown}{DN}$，$\therefore ∠COM = ∠COD = ∠DON$，$\therefore ∠COD = 20^{\circ}$，故選項 B 結(jié)論正確.$\because OE⊥CD$，$\therefore \overset{\frown}{CE} = \overset{\frown}{DE}$，$\therefore \overset{\frown}{ME} = \overset{\frown}{NE}$，$\therefore OE⊥MN$，$\therefore MN// CD$，故選項 C 結(jié)論正確.$\because \overset{\frown}{CM} = \overset{\frown}{CD} = \overset{\frown}{DN}$，$\therefore MC = CD = DN$，$\because MC + CD + DN ＞ MN$，$\therefore MN ＜ 3CD$，故選項 D 結(jié)論錯誤.故選 D.
15. [新課標 幾何直觀 2025 河南駐馬店西平期中]如圖,CD 是$\odot O$的直徑,點 A 是半圓上靠近點 D 的一個三等分點,B 是$\overset{\frown }{AD}$的中點,P 為直線 CD 上的一個動點,當$CD= 4$時,$AP+BP$的最小值為______.
答案：答案 $2\sqrt{2}$
解析 如圖，作點 A 關(guān)于 CD 的對稱點$A'$，連接$A'B$，交 CD 于點 P，此時$PA + PB$的值最小，最小值為$A'B$的長，連接$OA'$，OB，$\because$點 A 與$A'$關(guān)于 CD 對稱，點 A 是半圓上靠近點 D 的一個三等分點，$\therefore PA = PA'$，$∠A'OD = ∠AOD = 60^{\circ}$，$\because$點 B 是$\overset{\frown}{AD}$的中點，$\therefore ∠BOD = 30^{\circ}$，$\therefore ∠A'OB = ∠A'OD + ∠BOD = 90^{\circ}$，易知$OB = OA' = 2$，$\therefore A'B = 2\sqrt{2}$，$\therefore PA + PB = PA' + PB = A'B = 2\sqrt{2}$，$\therefore AP + BP$的最小值為$2\sqrt{2}$.
