22. 如图,⊙O的直径AB= 10,弦AC= 6,∠ACB的平分线交⊙O于点D. 求弦CD的长.

解：如图， 连 A D， B D
过点 作$ B E \perp C D $于 E
$\because A B $是直径
$\therefore \angle A C B=90^{\circ}，$$\angle A D B=90^{\circ}$
$\because A C=6，$ A B=10
$\therefore B C=\sqrt{A B^2-A C^2}$
$=\sqrt{10^2-6^2}=8$
$\because C D $平分$ \angle A C B$
$\therefore \angle B C D=45^{\circ}$
$\because B E \perp C D$
$\therefore C E=B E$
$\because C E^2+B E^2=B C^2，$ B C=8
$\therefore C E=B E=4 \sqrt{2}$
$\because C D $平分$ \angle A C B$
$\therefore\widehat{ A D}=\widehat{B D}$
$\therefore A D=B D$
$\because A D^2+B D^2=A B^2，$ A B=10
$\therefore B D=5 \sqrt{2}$
$\therefore $在$ R t \triangle B D E $中，
$\because B D=5 \sqrt{2}，$$ B E=4 \sqrt{2}$
$\therefore D E=\sqrt{B D^2-B E^2}$
$=\sqrt{(5 \sqrt{2})^2-(4 \sqrt{2})^2}=3 \sqrt{2}$
$\therefore C D=C E+D E$
$=4 \sqrt{2}+3 \sqrt{2}=7 \sqrt{2}$

23. 如图,AB是⊙O的直径,DF切⊙O于点D,BF⊥DF,AC//BF交BD的延长线于点C.
(1)∠ABC与∠C是否相等?为什么?
(2)设CA的延长线交⊙O于点E,BF交⊙O于点G,若$\overset{\frown}{DG}$为60°,试说明点D与点E关于直线AB对称.

解：(1)∠ ABC与$\angle C$相等
连接OD
$\because FD$切$\odot O$于点D
$\therefore OD\bot DF$
$\because BF\bot DF$
$\therefore OD//BF$
$\therefore \angle ODB=\angle FBD$
$\because OB=OD$
$\therefore \angle ODB=\angle OBD$
$\therefore \angle OBD=\angle FBD$
$\because AC//BF$
$\therefore \angle C=\angle FBD$
$\therefore \angle C=\angle OBD$
∴$\angle ABC=\angle C$
(2)连接BE，OE
$\because \widehat{DG}={60}^{\circ }$
$\therefore \angle GOD={60}^{\circ }$
$\therefore \angle FBD=\frac {1} {2}\angle GOD={30}^{\circ }$
$\because AC//BF$
$\therefore \angle C=\angle FBD={30}^{\circ } $
由 ( {1} )结论$\angle ABC=\angle C$得
$\angle ABC={30}^{\circ }$
$\because \angle BAE=\angle C+\angle ABC$
$\therefore \angle BAE={60}^{\circ }$
$\because OA=OE$
$\therefore \triangle OAE$是等边三角形
$\therefore \angle EOA={60}^{\circ }$
$\therefore \angle ABE=\frac {1} {2}\angle EOA={30}^{\circ }$
$\therefore \angle ABE=\angle ABC$
$\therefore \widehat{AD}=\widehat{AE}$
$\therefore $点D与点E关于直线AB对称