设AB=AD=1; ∠DAF=x
则DF=tan x FC=1-tan x BE=tan(45-x) EC=1-tan(45-x) ∠EFC=27+X
1-tan(45-x)=1-(1-tan x )/(1+ tan x )=2 tan x /(1+ tan x )
tan(27+x)= (1-tan(45-x))/(1-tan x)= 2 tan x /(1-tan x) (1+ tan x )
= 2 tan x /(1-tan^2 x)= tan2 x
∴ 27+x= 2 x
x=27
∠?=90-(45-x)=72(度)
请问您能分享一下您的做法吗?非常便捷?