题目内容
如图,点
在圆O上,
,
与
相交于点
,
,延长
到点
,使
,连结
.求证:直线
与圆O相切.![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230153337566399.jpg)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333584587.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333600531.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333616385.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333647398.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333647318.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333678680.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333694368.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333694302.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333709659.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333725395.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333725395.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230153337566399.jpg)
连OA,如图,
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230153337727195.png)
∵AE=
ED,FB=
BD,
∴AE:ED=FB:BD,
∴BE∥AF,
又∵AB=AC,
∴弧AB=弧AC,
∴OA⊥BC,
∴OA⊥AF,
∴直线AF与⊙O相切.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230153337727195.png)
∵AE=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333787338.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333787338.png)
∴AE:ED=FB:BD,
∴BE∥AF,
又∵AB=AC,
∴弧AB=弧AC,
∴OA⊥BC,
∴OA⊥AF,
∴直线AF与⊙O相切.
连OA,由AE=
ED,FB=
BD,则AE:ED=FB:BD,根据平行线分线段成比例定理得到BE∥AF;由AB=AC,根据垂径定理的推论得到OA⊥BC,则OA⊥AF,根据切线的判定定理即可得到结论.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333787338.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823015333787338.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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