Question

已知:如图,在△ABC中,∠ACB=90°,AC=BC,CD⊥AB,垂足是D,CE平分∠ACD,BF⊥CE,垂足是G,交

Answer 1

∵∠DHB=∠CHG,∠CHG+∠GCH=90,∠GCH+CED=90

∴∠DHB=∠CHG==∠CED

又∵△ABC中,∠ACB=90°AC=BC,CD⊥AB

∴CD=BD

∴ △CED≌△BHD

∴DH=DE,∠ECD=∠HBD=∠ACE (CE平分∠ACD)

过E作EM⊥AC于M

∵CE平分∠ACD

∴ EM=ED,而AE=√2EM

∴AE=√2DH

由∠HBD=∠ACE,∠A=∠A,知道△ACE∽△ABF

∴AF/AE=AB/AC=√2

∴AF=√2AE=√2(√2DH)=2DH