题目
题型:不详难度:来源:
(1)∠FEB=∠CEB;
(2)EF2=AD·BC.
![](http://img.shitiku.com.cn/uploads/allimg/20191108/20191108001056-69771.jpg)
答案
(2)见解析.
解析
由AB为⊙O的直径,得AE⊥EB,从而∠EAB+∠EBF=
![](http://img.shitiku.com.cn/uploads/allimg/20191108/20191108001056-75784.png)
又EF⊥AB,得∠FEB+∠EBF=
![](http://img.shitiku.com.cn/uploads/allimg/20191108/20191108001056-75784.png)
故∠FEB=∠CEB.
(2)由BC⊥CE,EF⊥AB,∠FEB=∠CEB,BE是公共边,
得Rt△BCE≌Rt△BFE,所以BC=BF.
类似可证:Rt△ADE≌Rt△AFE,得AD=AF.
又在Rt△AEB中,EF⊥AB,故EF2=AF·BF,
所以EF2=AD·BC.
核心考点
试题【如图,AB为⊙O的直径,直线CD与⊙O相切于E,AD垂直CD于D,BC垂直CD于C,EF垂直AB于F,连结AE,BE.证明:(1)∠FEB=∠CEB;(2)EF】;主要考察你对圆锥曲线性质探讨等知识点的理解。[详细]
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A.直角三角形 |
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