过D作AE的垂线,交AE、AB于F、G,过G做GH⊥BC交BC于G
设AE=BC=x,GF=a
可知GD∥BC,GH∥AE∥CD,故FD=CE=x-5;GH=EF=CD=8,故AF=x-8
故∠GDB=∠CBD,因BD等分∠ABC,故∠GDB=∠GBD,故GB=GD=a+x-5
因GF∥BE,可知△AFG∽△ABE,FG/EB=AF/AE,即a/5=(x-8)/x,故5x-ax=40
因GH∥AE,故△BGH∽△BAE,故GH/AE=BG/BA,即8/x=(a+x-5)/√(x^2+25)
即8√(x^2+25)=ax+x^2-5x=x^2-40
64x^2+1600=x^4-80x^2+1600
x^4=144x^2
因x>0
故x^2=144
x=12
可知AF=x-8=4,FD=x-5=7
故AD=√(AF^2+FD^2)=√65![](https://imgsa.baidu.com/forum/w%3D580/sign=d7030131963df8dca63d8f99fd1372bf/cc009313b07eca80c2f43621982397dda0448310.jpg)
设AE=BC=x,GF=a
可知GD∥BC,GH∥AE∥CD,故FD=CE=x-5;GH=EF=CD=8,故AF=x-8
故∠GDB=∠CBD,因BD等分∠ABC,故∠GDB=∠GBD,故GB=GD=a+x-5
因GF∥BE,可知△AFG∽△ABE,FG/EB=AF/AE,即a/5=(x-8)/x,故5x-ax=40
因GH∥AE,故△BGH∽△BAE,故GH/AE=BG/BA,即8/x=(a+x-5)/√(x^2+25)
即8√(x^2+25)=ax+x^2-5x=x^2-40
64x^2+1600=x^4-80x^2+1600
x^4=144x^2
因x>0
故x^2=144
x=12
可知AF=x-8=4,FD=x-5=7
故AD=√(AF^2+FD^2)=√65
![](https://imgsa.baidu.com/forum/w%3D580/sign=d7030131963df8dca63d8f99fd1372bf/cc009313b07eca80c2f43621982397dda0448310.jpg)