如图已知三角形ABC中,∠ACB=90°,AC=BC,AE是BC边的中线,CD⊥AE于点D,CD的延长线J交AB于点F,

证明：过点B作BG⊥BC交CF的延长线于点G∵∠ACB＝90,AC＝BC∴∠CAE+∠AEC＝90,∠CAB＝∠CBA＝45∵CD⊥AE∴∠BCG+∠AEC＝90∴∠CAE＝∠BCG∵BG⊥BC∴∠CBG＝∠ACB＝90∴△ACE≌△CBG （ASA）∴CE＝BG,∠AEC＝∠G∵AE是...