(2)如图(2),延长BC、FE交于点P ∵正方形ABCD, ∴AD∥BC, ∴△DEF∽△CEP, ∵E为CD的中点, ∴,PF=2EF, ∵∠BFE=∠FBC, ∴PB=PF, ∵AF=a, ∴PC=DF=4-a,PB=PF=8-a,EF=PF/2=8-a/2, ∵Rt△DEF中,EF2=DE2+DF2, ∴(8-a/2)2=22+(4-a)2, 整理,得3a2-16a+16=0 , 解得a1=4/3,a2=4, ∵F点不与D点重合, ∴a=4不成立,a=4/3,tan∠AFB=AB/AF=3; | |