Question

△ABC中,∠ACB=90°AD⊥AB,AD=AB,BE⊥DC于E,AF⊥AC交EB的延长线于F.求证AF=AC

Answer 1

∵AD⊥AB,BE⊥DC,AF⊥AC

∴∠DAB=∠CEB=∠CAE=∠ACB=90º

∵∠D=90º-∠AOD

∠ABF=90º-∠BOE

∠AOD=∠BOE

∴∠D=∠ABF

∵∠DAC=90º-∠CAB

∠BAF =90º-∠CAB

∴∠DAC =∠BAF

又∵AD=AB

∴⊿DAC≌⊿BAF（AAS）

∴AC=AF