23. 如图，正六边形ABCDEF内接于$\odot O$，半径为2 cm.
（1）求CD的长度；
（2）若G为CD的中点，连接AG，求AG的长度.

23. (1) 如图①,连接OC,OD,$\because$六边形ABCDEF是正六边形,$\therefore ∠COD=\frac{360°}{6}=60°$,又OC,OD是$\odot O$的半径,且半径为2 cm,$\therefore OC=OD=2$ cm,$\therefore △COD$是等边三角形,$\therefore CD=OC=2$ cm.
(2) 如图②,连接AC,AD,则AD为$\odot O$的直径,$\therefore AD=4$ cm,$∠ACD=90°$,由(1)得$CD=2$ cm.在$Rt△ACD$中,$∠ACD=90°$,$\therefore AC=\sqrt{AD^2-CD^2}=\sqrt{4^2-2^2}=2\sqrt{3}$ cm.$\because G$为CD的中点,$\therefore CG=\frac{1}{2}CD=1$ cm.在$Rt△ACG$中,$∠ACG=90°$,$\therefore AG=\sqrt{AC^2+CG^2}=\sqrt{(2\sqrt{3})^2+1^2}=\sqrt{13}$ cm.

24. 如图，BE是$\odot O$的直径，点A和点D是$\odot O$上的两点，连接AE，AD，DE，过点A作射线交BE的延长线于点C，使$∠ EAC=∠ EDA$。
（1）求证：AC是$\odot O$的切线；
（2）若$CE=AE=2\sqrt{3}$，求阴影部分的面积.

24. (1) 证明:如图,连接OA,过O作$OF⊥AE$于点F,$\therefore ∠AFO=90°$,$\therefore ∠EAO+∠AOF=90°$.$\because OA=OE$,$\therefore ∠EOF=∠AOF=\frac{1}{2}∠AOE$.$\because ∠EDA=\frac{1}{2}∠AOE$,$\therefore ∠EDA=∠AOF$.$\because ∠EAC=∠EDA$,$\therefore ∠EAC=∠AOF$,$\therefore ∠EAO+∠EAC=90°$.$\because ∠EAC+∠EAO=∠CAO$,$\therefore ∠CAO=90°$,$\therefore OA⊥AC$,$\therefore AC$是$\odot O$的切线.
(2) $\because CE=AE=2\sqrt{3}$,$\therefore ∠C=∠EAC$.$\because ∠EAC+∠C=∠AEO$,$\therefore ∠AEO=2∠EAC$.$\because OA=OE$,$∠AEO=∠EAO$,$\therefore ∠EAO=2∠EAC$.$\because ∠EAO+∠EAC=90°$,$\therefore ∠EAC=30°$,$∠EAO=60°$,$\therefore △OAE$是等边三角形,$\therefore ∠EOA=60°$,$\therefore OA=AE=2\sqrt{3}$,$\therefore S_{扇形AOE}=\frac{60· π×(2\sqrt{3})^2}{360}=2π$.在$Rt△OAF$中,$OF=OA· \sin∠EAO=2\sqrt{3}×\frac{\sqrt{3}}{2}=3$,$\therefore S_{△AOE}=\frac{1}{2}AE· OF=\frac{1}{2}×2\sqrt{3}×3=3\sqrt{3}$,$\therefore$阴影部分的面积$=2π-3\sqrt{3}$.