本题考查圆的相关内容。如相切等。本题利用等腰三角形的性质证明Rt△ACE≌Rt△BCF然后利用相似和全等求解相关问题。
(1)猜想:OG⊥CD.
证明:如图,连结OC、OD,∵OC=OD,G是CD的中点,
∴由等腰三角形的性质,有CG⊥CD. (3分)
(2)证明: ∵AB是⊙O的直径, ∴∠ACB=90°.
在Rt△ACE和Rt△BCF中
∠CAE=∠CBF, ∠ACE=∠BCF=90°,AC=BC.
∴Rt△ACE≌Rt△BCF
∴AE="BF." (7分)
(3)解:过点O作BD的垂线,垂足为H.则H为BD的中点.
∴OH=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346859313.png)
AD,即AD=2OH.
又∠CAD=∠BAD ,∴CD="BD," ∴OH=OG.
在Rt△BDE和Rt△ADB中,∠DBE=∠DAC=∠BAD,
∴Rt△BDE∽Rt△ADB, ∴BD=AD·DE=2OG·DE=6(2-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
)
又BD="FD," ∴BF="2BD." ∴BF=4BD=24(2-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
).……①
设AC=x,则BC=x,AB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
x.
∵AD是∠BAC的平分线,∴∠FAD=∠BAD.
在Rt△ABD和Rt△AFD中,∠ADB=∠ADF=90°,AD=AD,∠FAD=∠BAD,
∴Rt△ABD≌Rt△AFD.∴AF=AB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
x-x=(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
-1)x
在Rt△BCF中BF=BC+CF=x+[(
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
-1)x] =2(2-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
)x……②
由①、②解得x=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346984307.png)
或-2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346984307.png)
(舍去).
∴AB=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
x=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346828310.png)
·2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013346984307.png)
=2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013347062334.png)
.
∴S
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013347062320.png)
=π·(2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823013347062334.png)
)=6π