题目内容
如图
,
于
,
交
于点
,
,则
___________________
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044105842.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044120557.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044136303.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044136520.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044152382.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044198289.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044214533.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044230398.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044245493.png)
∵BD与AH交于点P,则由AC=AD,AH⊥CD得∠ACH=∠ADH.
又AB=AD,故∠ADB=∠ABD.
从而,∠ABP=∠ACP.可知A、B、C、P四点共圆.
∵∠APC=90°+∠PCH=∠BCD,∠CBP=∠CAP,
∴△APC∽△BCD.
∴AC•BC=AP•BD.
∴BD=AC•
,
∴BD=![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044276504.png)
又AB=AD,故∠ADB=∠ABD.
从而,∠ABP=∠ACP.可知A、B、C、P四点共圆.
∵∠APC=90°+∠PCH=∠BCD,∠CBP=∠CAP,
∴△APC∽△BCD.
∴AC•BC=AP•BD.
∴BD=AC•
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044261534.png)
∴BD=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013044276504.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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