题目内容
如图,射线AM,BN都垂直于线段AB,点E为AM上一点,过点A作BE的垂线AC分别交BE,BN于点F,C,过点C作AM的垂线CD,垂足为D.若CD=CF,则![](http://thumb.1010pic.com/pic1/imagenew2/czsx/2/67502.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/3/67503.gif)
:
解析:
见题图,设
.
因为Rt△AFB∽Rt△ABC,所以
.
又因为 FC=DC=AB,所以
即
,
解得
,或
(舍去).
又Rt△
∽Rt△
,所以![](http://thumb.1010pic.com/pic1/imagenew2/czsx/13/67513.gif)
, 即
=
.
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/4/67504.gif)
见题图,设
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/5/67505.gif)
因为Rt△AFB∽Rt△ABC,所以
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/6/67506.gif)
又因为 FC=DC=AB,所以
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/7/67507.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/8/67508.gif)
解得
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/9/67509.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/10/67510.gif)
又Rt△
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/11/67511.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/12/67512.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/13/67513.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/14/67514.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/15/67515.gif)
![](http://thumb.1010pic.com/pic1/imagenew2/czsx/14/67514.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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