Question

求大神解一道挺难的几何题？

如图

Answer 1

PF,PE为角平分线

设BP=x,AC=y

AP=4-x

CP/PA=CF/FA

CF=y/(5-x)

AF=(4-x)y/(5-x)

∵△APF∽△ACP

∴(4-x)^2=(4-x)y^2/(5-x)

y^2=(5-x)(4-x)

BC=√(16-y^2)

BP/PC=BE/EC

BE=x√(16-y^2)/(x+1)

∵△BPE∽△BCP

∴x^2=x(16-y^2)/(x+1)

x(x+1)=16-y^2

y^2=-x^2-x+16

-x^2-x+16=x^2-9x+20

2x^2-8x+4=0

x^2-4x+2=0

x=2±√2

∵∠BAC<45°

∴x=2-√2

AP/BP=(2+√2)/(2-√2)=3+2√2

有点麻烦

望采纳