很简单啊要不要过程
(1)①BF=AD,BF⊥AD;②BF=AD,BF⊥AD仍然成立,证明:∵△ABC是等腰直角三角形,∠ACB=90°,∴AC=BC,∵四边形CDEF是正方形,∴CD=CF,∠FCD=90°,∴∠ACB+∠ACF=∠FCD+∠ACF,即∠BCF=∠ACD,在△BCF和△ACD中BC=AC∠BCF=∠ACDCF=CD∴△BCF≌△ACD(SAS),∴BF=AD,∠CBF=∠CAD,又∵∠BHC=∠AHO,∠CBH+∠BHC=90°,∴∠CAD+∠AHO=90°,∴∠AOH=90°,∴BF⊥AD;(2)连接DF, ∵四边形CDEF是矩形,∴∠FCD=90°,又∵∠ACB=90°,∴∠ACB=∠FCD∴∠ACB+∠ACF=∠FCD+∠ACF,即∠BCF=∠ACD,∵AC=4,BC=3,CD= 43,CF=1,∴ BCAC= CFCD= 43∴△BCF∽△ACD,∴∠CBF=∠CAD,又∵∠BHC=∠AHO,∠CBH+∠BHC=90°∴∠CAD+∠AHO=90°,∴∠AOH=90°,∴BF⊥AD,∴∠BOD=∠AOB=90°,∴BD2=OB2+OD2,AF2=OA2+OF2,AB2=OA2+OB2,DF2=OF2+OD2,∴BD2+AF2=OB2+OD2+OA2+OF2=AB2+DF2,∵在Rt△ABC中,∠ACB=90°,AC=4,BC=3,∴AB2=AC2+BC2=32+42=25,∵在Rt△FCD中,∠FCD=90°,CD= 43,CF=1,∴DF2=CD2+CF2=( 43)212= 259∴BD2+AF2=AB2+DF2=25+ 259= 2509故答案为:解:(1)①BF=AD,BF⊥AD;②BF=AD,BF⊥AD仍然成立,证明:∵△ABC是等腰直角三角形,∠ACB=90°,∴AC=BC,∵四边形CDEF是正方形,∴CD=CF,∠FCD=90°,∴∠ACB+∠ACF=∠FCD+∠ACF,即∠BCF=∠ACD,在△BCF和△ACD中BC=AC∠BCF=∠ACDCF=CD∴△BCF≌△ACD(SAS),∴BF=AD,∠CBF=∠CAD,又∵∠BHC=∠AHO,∠CBH+∠BHC=90°,∴∠CAD+∠AHO=90°,∴∠AOH=90°,∴BF⊥AD;(2)连接DF, ∵四边形CDEF是矩形,∴∠FCD=90°,又∵∠ACB=90°,∴∠ACB=∠FCD∴∠ACB+∠ACF=∠FCD+∠ACF,即∠BCF=∠ACD,∵AC=4,BC=3,CD= 43,CF=1,∴ BCAC= CFCD= 43∴△BCF∽△ACD,∴∠CBF=∠CAD,又∵∠BHC=∠AHO,∠CBH+∠BHC=90°∴∠CAD+∠AHO=90°,∴∠AOH=90°,∴BF⊥AD,∴∠BOD=∠AOB=90°,∴BD2=OB2+OD2,AF2=OA2+OF2,AB2=OA2+OB2,DF2=OF2+OD2,∴BD2+AF2=OB2+OD2+OA2+OF2=AB2+DF2,∵在Rt△ABC中,∠ACB=90°,AC=4,BC=3,∴AB2=AC2+BC2=32+42=25,∵在Rt△FCD中,∠FCD=90°,CD= 43,CF=1,∴DF2=CD2+CF2=( 43)212= 259∴BD2+AF2=AB2+DF2=25+ 259= 2509
追问当然要自己写的