Question

∠BAC=90°，AB=AC,D点在AC上，E点在BA的延长线上，BD=CE,BD的延长线交CE于F，试证明：BF⊥CE

Answer 1

△ABD和△AEC中，

∠BAD=∠EAC=90º

AB=AC,BD=EC

所以 △BAD≌△EAC

故：∠ABD=∠ACE 即：∠ABD=∠FCD ,又因∠FDC=∠ADB

所以：∠BAD=∠DFC=90º 即：DF⊥EC ∴BF⊥EC

Answer 2

AB=AC ∠BAC=∠CAE=90° BD=CE
△BDE≌△CEA
∠BDA=∠E
∠BDA+∠EBF=90°
∴∠EBF+∠E=90°
∴BF⊥CE

Answer 3

∵BA＝CA,∠BAC＝∠EAC,BD＝CE,∴△ABD≌△EAC ∴∠ABD＝∠DCF 又∵∠ADB＝∠FDC

∴∠BAD＝∠CFA 即BF⊥CE

照顾点呗~