题目内容
如图,△ABC中,CD⊥AB于D,E为BC中点,延长AC、DE相交于点F,
求证
=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345150549.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230123451662445.png)
求证
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345135543.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345150549.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230123451662445.png)
方法一:作FG∥BC交AB延长线于点G.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230123451823353.png)
∵ BC∥GF,
∴
=
.
又 ∠BDC=90°,BE=EC,
∴ BE=DE.
∵ BE∥GF,
∴
=
=1.
∴ DF=GF.
∴
=
.
方法二:作EH∥AB交AC于点H.
∵
=
,
=
,
∠BDC=90°,BE=EC,
∴ BE=DE.
∴
=
.
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230123451823353.png)
∵ BC∥GF,
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345135543.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345213547.png)
又 ∠BDC=90°,BE=EC,
∴ BE=DE.
∵ BE∥GF,
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345228549.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345244543.png)
∴ DF=GF.
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345135543.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345150549.png)
方法二:作EH∥AB交AC于点H.
∵
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345135543.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345306575.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345150549.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345338583.png)
∠BDC=90°,BE=EC,
∴ BE=DE.
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345135543.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012345150549.png)
略
![](http://thumb.zyjl.cn/images/loading.gif)
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