9. 如图,已知在半圆$\widehat{ADB}$中,$AD=DC$,$∠ CAB=30°$,$AC=2\sqrt{3}$,求$AD$的长度.

9. 解:连接 OD,交 AC 于点 H. $\because AD=DC, \therefore \overset{\frown}{AD}=\overset{\frown}{CD}, \therefore OD⊥ AC. \because AC=2\sqrt{3}, \therefore AH=CH=\sqrt{3}.$ 又$\because ∠ OAH=30°, \therefore OH=1, OA=2, ∠ AOH = 60°. \because OA = OD, \therefore △ OAD$是等边三角形，$\therefore AD=OA=2.$

10. 如图，以$□ ABCD$的顶点A为圆心，AB长为半径作圆，交AD，BC于点E，F，延长BA交$\odot A$于点G. 求证：$\overset{\frown}{GE}=\overset{\frown}{EF}$.

10. 证明:连接 AF. $\because$ 四边形 ABCD 为平行四边形, $\therefore AD// BC, \therefore ∠ B=∠ GAE, ∠ EAF=∠ AFB.$ 又$\because AB=AF, \therefore ∠ B=∠ AFB, \therefore ∠ GAE=∠ EAF, \therefore \overset{\frown}{GE}=\overset{\frown}{EF}.$

11.(核心素养·推理能力)如图,在$\odot O$中,点C是优弧ACB的中点,D,E分别是OA,OB上的点,且$AD=BE$,弦CM,CN分别过点D,E.求证:
(1)$CD=CE$;
(2)$\overset{\frown}{AM}=\overset{\frown}{BN}$.

11. (1)证明:连接 OC. $\because$ 点 C 是优弧 ACB 的中点，$\therefore \overset{\frown}{AC}=\overset{\frown}{BC}, \therefore ∠ COD=∠ COE. \because OA=OB, AD=BE, \therefore OD=OE. \because OC=OC, \therefore △ COD≌△ COE(\mathrm{SAS}). \therefore CD=CE.$ (2)连接 OM,ON. $\because △ COD≌△ COE, \therefore ∠ CDO=∠ CEO, ∠ OCD=∠ OCE. \because OC=OM=ON, \therefore ∠ OCM=∠ OMC, ∠ OCN=∠ ONC. \therefore ∠ OMD=∠ ONE. \because ∠ ODC = ∠ DMO + ∠ MOD, ∠ CEO = ∠ ENO + ∠ EON, \therefore ∠ MOD=∠ NOE. \therefore \overset{\frown}{AM}=\overset{\frown}{BN}.$