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22. (12分)如图,已知$AB$是$\odot O$的直径,点$C$在$\odot O$上,延长$BC$至点$D$,使得$DC=BC$,直线$DA$与$\odot O$的另一个交点为$E$,连接$AC,CE$.
(1)求证:$CD=CE$;
(2)若$AC=2$,$∠ E=30^{\circ}$,求阴影部分(弓形)面积.
(1)求证:$CD=CE$;
(2)若$AC=2$,$∠ E=30^{\circ}$,求阴影部分(弓形)面积.
答案
22. (1)证明:$\because AB$是$\odot O$的直径,
$\therefore ∠ACB=90^{\circ }$.
$\because DC=BC$,
$\therefore AD=AB$.
$\therefore ∠D=∠ABC$.
$\because ∠E=∠ABC$,
$\therefore ∠E=∠D$.
$\therefore CD=CE$.
(2)解:由(1)可知$∠ABC=∠E=30^{\circ },∠ACB=90^{\circ }$,
$\therefore ∠CAB=60^{\circ },AB=2AC=4$.
在$Rt△ABC$中,由勾股定理得到$BC=2\sqrt {3}$.
连接$OC$,则$∠COB=120^{\circ }$.
$\therefore S_{阴影}=S_{扇形OBC}-S_{△OBC}=\frac {120×π×2^{2}}{360}-\frac {1}{2}×\frac {1}{2}×2\sqrt {3}×2=\frac {4π}{3}-\sqrt {3}$.
$\therefore ∠ACB=90^{\circ }$.
$\because DC=BC$,
$\therefore AD=AB$.
$\therefore ∠D=∠ABC$.
$\because ∠E=∠ABC$,
$\therefore ∠E=∠D$.
$\therefore CD=CE$.
(2)解:由(1)可知$∠ABC=∠E=30^{\circ },∠ACB=90^{\circ }$,
$\therefore ∠CAB=60^{\circ },AB=2AC=4$.
在$Rt△ABC$中,由勾股定理得到$BC=2\sqrt {3}$.
连接$OC$,则$∠COB=120^{\circ }$.
$\therefore S_{阴影}=S_{扇形OBC}-S_{△OBC}=\frac {120×π×2^{2}}{360}-\frac {1}{2}×\frac {1}{2}×2\sqrt {3}×2=\frac {4π}{3}-\sqrt {3}$.
