22. 如图，AB是$\odot O$的直径，弦$CD⊥ AB$于点$E$，点$P$在$\odot O$上，弦$PB$与$CD$交于点$F$，且$FP=FD$。
（1）求证：$PD// CB$；
（2）若$AB=26$，$CD=24$，求$BE$的长度。

22. （1）证明:$\because FP=FD$,$\therefore ∠P=∠FDP.\because ∠P=∠C,\therefore ∠C=∠FDP,\therefore PD// CB.$
（2）如图,连接OC,$\because AB$是直径,$AB⊥ CD,\therefore E$为CD的中点.$\because AB=26,CD=24,\therefore OC=OB=13,CE=12,\therefore OE=\sqrt{OC^2-CE^2}=\sqrt{13^2-12^2}=5,\therefore BE=OB-OE=13-5=8.$

23. 如图，AB是$\odot O$的直径，点C在$\odot O$上，作$CG ⊥ AB$于点D，交$\odot O$于点G，$∠ ACG$的平分线交AB于点E，交$\odot O$于点F，连接AF，BF.
（1）若$\odot O$的半径为6，$AD=4$，求弦CG的长；
（2）求证：$AF=EF$.

23. （1）如图,连接OC,$\because CG⊥ AB$,AB是$\odot O$的直径,$\therefore CD=GD=\frac{1}{2}CG,∠ODC=90°.\because AD=4,\odot O$的半径为6,$\therefore OD=OA-AD=2.$在$\mathrm{Rt}△ OCD$中,$CD=\sqrt{OC^2-OD^2}=\sqrt{6^2-2^2}=4\sqrt{2},\therefore CG=2CD=8\sqrt{2}.$
（2）证明:$\because CG⊥ AB,\therefore ∠ADC=90°,\therefore ∠CED=90°-∠GCF.\because ∠AEF=∠CED,\therefore ∠AEF=90°-∠GCF.\because AB$是$\odot O$的直径,$\therefore ∠AFB=90°,\therefore ∠FAE=90°-∠ABF.\because \overset{\frown}{AF}=\overset{\frown}{AF},\therefore ∠ABF=∠ACF.\because CF$平分$∠ACG,\therefore ∠ACF=∠GCF=∠ABF,\therefore ∠AEF=∠FAE,\therefore AF=EF.$

24. 如图，已知AB是半圆的直径，点C在半圆上，$CE ⊥ AB$，垂足为E，点D是$\overset{\frown}{BC}$的中点，AD交CE于点F，交BC于点G.
(1) $∠ CAD = ∠$$= ∠$；
(2) 判断$△ FGC$的形状，并说明理由；
(3) 若$\overset{\frown}{CD}$为$60°$，$AB=16$，求阴影部分的面积.

24. (1) $\because \overset{\frown}{CD}=\overset{\frown}{CD},\therefore ∠CAD=∠CBD.\because D$为$\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{CD}=\overset{\frown}{BD},\therefore ∠CAD=∠BAD.$综上所述,$∠CAD=∠BAD=∠CBD.$
（2）如图,$△ FGC$是等腰三角形,理由:$\because AB$是$\odot O$的直径,$\therefore ∠ACB=90°,\therefore ∠CAD+∠AGC=90°.\because CE⊥ AB,\because ∠AFE+∠BAD=90°,$D为$\overset{\frown}{BC}$的中点,$\therefore \overset{\frown}{CD}=\overset{\frown}{BD},\therefore ∠CAD=∠BAD,\therefore ∠AGC=∠AFE.\because ∠AFE=∠CFG,\therefore ∠CGF=∠CFG,\therefore CF=CG,\therefore △ CFG$是等腰三角形.
（3）如图,连接OC,OD,$\because \overset{\frown}{CD}$为$60°,\therefore ∠COD=60°.\because D$为$\overset{\frown}{BC}$的中点,$\therefore ∠BOD=∠COD=60°,\therefore ∠AOC=180°-60°-60°=60°.\because OA=OC,\therefore △ AOC$为等边三角形,$\therefore OA=AC=OC=\frac{1}{2}AB=8.\because CE⊥ AB,\therefore AE=\frac{1}{2}OA=4,\therefore CE=\sqrt{AC^2-AE^2}=\sqrt{8^2-4^2}=4\sqrt{3}.\therefore S_{\mathrm{阴影}}=S_{\mathrm{扇形}AOC}-S_{△ AOC}=\frac{60×π×8^2}{360}-\frac{1}{2}×8×4\sqrt{3}=\frac{32}{3}π-16\sqrt{3}.$