9. 易错题 一题多解 [2025徐州丰县期中]在$\odot O$中，$\overset{\frown}{AB}=2\overset{\frown}{CD}$，则弦AB与弦CD的大小关系是（）

A.$AB＞2CD$
B.$AB=2CD$
C.$AB＜2CD$
D.$AB=CD$

9 C 通解 如图,取$\overset{\frown}{AB}$的中点 E,连接 AE,BE,则$\overset{\frown}{AE}=\overset{\frown}{BE}.\because \overset{\frown}{AB}=2\overset{\frown}{CD},\therefore \overset{\frown}{AE}=\overset{\frown}{BE}=\overset{\frown}{CD},\therefore AE = BE = CD.\because AE + BE ＞ AB,\therefore 2CD ＞ AB.$
另解 在$\overset{\frown}{AC}$上取一点 E,使$CE = CD$,连接 DE,则$\overset{\frown}{CD}=\overset{\frown}{CE}$,$\overset{\frown}{DCE}=2\overset{\frown}{CD},\because \overset{\frown}{AB}=2\overset{\frown}{CD},\therefore \overset{\frown}{AB}=\overset{\frown}{DCE},\therefore AB = DE.\because CE + CD ＞ DE,\therefore 2CD ＞ AB.$

避坑指南
本题的易错之处是由$\overset{\frown}{AB}=2\overset{\frown}{CD}$直接得到$AB=2CD$.
在同圆或等圆中,由弧相等可推出相应的弦相等,但当弧有倍数关系时,弦没有相应的倍数关系.

10. 在半径为1的圆中，长度等于$\sqrt{2}$的弦所对的弧的度数为（）

A.$90°$
B.$145°$
C.$90°$或$270°$
D.$270°$或$145°$

10 C 由题意,知半径$r=1$,弦长为$\sqrt{2}$,$\because (\sqrt{2})^2 = 1^2 + 1^2$,$\therefore$ 根据勾股定理的逆定理,得长度等于$\sqrt{2}$的弦与两条半径可构成直角三角形. 又$\because$ 弦所对的弧有优弧、劣弧,$\therefore$ 长度等于$\sqrt{2}$的弦所对弧的度数为$90°$或$270°$.

11. 如图,在扇形OAB中,∠AOB = 110°,将扇形OAB沿过点B的直线折叠,点O恰好落在$\overset{\frown}{AB}$上的点D处,折痕交OA于点C,则$\overset{\frown}{AD}$的度数为。

11 50° 如图,连接 OD. 由折叠的性质,知$OB = BD,\therefore OB = BD = OD$,$\therefore △ OBD$是等边三角形,$\therefore ∠BOD = ∠OBD = ∠ODB = 60°$,$\therefore ∠AOD = ∠AOB - ∠BOD = 110° - 60° = 50°$,即$\overset{\frown}{AD}$的度数为50°.

12. 如图,已知AB是$\odot O$的直径,M,N分别是AO,BO的中点,$CM ⊥ AB,DN ⊥ AB$.求证:$\overset{\frown}{AC}=\overset{\frown}{BD}$.

12 证明:如图,连接 OC,OD.
$\because AB$ 是$\odot O$的直径,M,N 分别是 AO,BO 的中点,
$\therefore OM = ON.$
$\because CM ⊥ AB,DN ⊥ AB,$
$\therefore ∠OMC = ∠OND = 90°.$
在$\mathrm{Rt}△ OMC$和$\mathrm{Rt}△ OND$中,
$\begin{cases}OM = ON,\\OC = OD,\end{cases}$
$\therefore \mathrm{Rt}△ OMC ≌ \mathrm{Rt}△ OND,$
$\therefore ∠COM = ∠DON,\therefore \overset{\frown}{AC}=\overset{\frown}{BD}.$

13. [2026南京中华附初月考]如图,已知AB为$\odot O$的直径,$∠ DOC=90°$,$∠ DOC$在直径AB上方绕点O旋转,D,C两点不与A,B重合.
(1)求证:$\overset{\frown}{AD} + \overset{\frown}{BC} = \overset{\frown}{CD}$.
(2)$AD + BC = CD$成立吗?为什么?

13 (1)证明:$\because AB$ 为$\odot O$的直径,$∠DOC = 90°$,
$\therefore ∠AOD + ∠BOC = ∠DOC = 90°.$
$\therefore \overset{\frown}{AD}+\overset{\frown}{BC}=\overset{\frown}{CD}.$
(2)解:不成立. 理由如下:
如图,在$\overset{\frown}{CD}$上截取$\overset{\frown}{DE}=\overset{\frown}{AD}$,连接 EC,DE,则$DE = AD.$
由(1)知$\overset{\frown}{AD}+\overset{\frown}{BC}=\overset{\frown}{CD}$,即$\overset{\frown}{AD}+\overset{\frown}{BC}=\overset{\frown}{DE}+\overset{\frown}{EC}$,
又$\because \overset{\frown}{DE}=\overset{\frown}{AD},\therefore \overset{\frown}{BC}=\overset{\frown}{EC},\therefore BC = EC.$
在$△ DEC$中,$\because DE + EC ＞ DC,\therefore AD + BC ＞ CD.$

14. [2026 盐城建湖汇杰中学月考] 如图，OA, OB 是$\odot O$的半径，且$\overset{\frown}{AC} = \overset{\frown}{BC}$，弦 CM, CN 分别经过 OA, OB 的中点 D,E.
求证:(1)$CD = CE$；
(2)$CM = CN$.

14 证明:(1)$\because OA,OB$ 是$\odot O$的半径,$\therefore OA = OB.$
$\because D,E$ 分别是 OA,OB 的中点,
$\therefore OD = \frac{1}{2}OA,OE = \frac{1}{2}OB,\therefore OD = OE.$
$\because \overset{\frown}{AC}=\overset{\frown}{BC},\therefore ∠AOC = ∠BOC.$
在$△ DOC$和$△ EOC$中,$\begin{cases}OD = OE,\\∠DOC = ∠EOC,\\OC = OC,\end{cases}$
$\therefore △ DOC ≌ △ EOC,\therefore CD = CE.$
(2)如图,连接 OM,ON.

由(1),得$△ DOC ≌ △ EOC$,
$\therefore ∠OCD = ∠OCE$,即$∠OCM = ∠OCN.$
$\because OM = OC,ON = OC,$
$\therefore ∠OMC = ∠OCM,∠ONC = ∠OCN,$
$\therefore ∠MOC = ∠NOC,\therefore CM = CN.$