解:(1)①22.5°…………………………2分
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093557-17524.jpg) 证明:如图1,过点D作DG∥CA,与BE的延长线相交于点G,与AB相交于点H
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093558-40083.jpg) 则∠GDB=∠C ∠BHD=∠A=90°=∠GHB
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093558-92813.jpg) 又∵DE=DE,∠DEB=∠DEG=90° ∴△DEB≌△DEG
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093558-74960.jpg) ∵AB=AC ∠A=90° ∴∠ABC=∠C=∠GDB ∴HB=HD ∵∠DEB=∠BHD=90° ∠BFE=∠DFH ∴∠EBF=∠HDF ∴△GBH≌△FDH ∴GB=FD…………………………6分
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093559-33516.jpg) (2)如图1,过点D作DG∥CA,与BE的延长线相交于点G,与AB相交于点H
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093559-31196.jpg)
又∵DG∥CA ∴△BHD∽△BAC
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093600-43120.jpg) 第二种解法: 解:(1)①∵AB=AC∠A=90° ∴∠ABC=∠C=45° ∵∠EDB= ∠C ∴∠EDB=22.5° ∵BE⊥DE ∴∠EBD=67.5° ∴∠EBF=67.5°-45°=22.5° ②在△BEF和△DEB中 ∵∠E=∠E=90° ∠EBF=∠EDB=22.5° ∴△BEF∽△DEB 如图:BG平分∠ABC,
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093600-66434.jpg) ∴BG=GD△BEG是等腰直角三角形 设EF=x,BE=y, 则:BG=GD= y FD= y+y-x ∵△BEF∽△DEB ∴ =![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093601-79503.gif) 即: =![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093601-41161.gif) 得:x=( -1)y ∴FD= y+y-( -1)y=2y ∴FD=2BE. (2)如图:作∠ACB的平分线CG,交AB于点G,
![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093602-94014.jpg) ∵AB=kAC ∴设AC=b,AB=kb,BC= b 利用角平分线的性质有: =![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093602-72047.gif) 即: =![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093603-36771.gif) 得:AG=![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093603-70937.gif) ∵∠EDB= ∠ACB ∴tan∠EDB=tan∠ACG=![](http://img.shitiku.com.cn/uploads/allimg/20191103/20191103093603-95248.gif) ∵∠EDB= ∠ACB ∠ABC=90°-∠ACB ∴∠EBF=90°-∠ABC-∠EDB= ∠ACB ∴△BEF∽△DEB ∴EF= BE ED= BE=EF+FD ∴FD= BE- BE= BE. ∴ = . |