(1)∠BCE=90°-∠CAD=∠BAF,∠CBE=∠ABF∴△ABF∽△CBE∴AF:CE=AB:CB由角平分线定理,AB:CB=AE:CE∴AF:CE=AE:CE∴AF=AE∴△AEF为等腰三角形,H为EF中点∴AH⊥EF(2)C△AHF=AF+FH+AH,C△BDE=BD+DE+BE,C△BAF=AB+BF+AF设BF的长为单位1,AF/BF=x,则,AF=x。又设∠CBE=θ,则,BD=cosθ, FH=AF*sinθ=x*sinθ, AH=x*cosθ,BE=BF+2FH=1+2x*sinθ,DE=√(BD^2+BE^2-2BD*BE*cosθ),AB=√(... |