Question

已知△ABC中，∠BAC=90°，AB=AC,AE是过点A的直线,BD⊥AE于点D,CE⊥AE，求证BD=CE+DE

Answer 1

∠BAC=90°，BD⊥AE，
∴∠ABD=∠CAE,
又AB=AC,CE⊥AE,
∴△ABD≌△CAE(ASA),
∴BD=AE,AD=CE,
∴CE+DE=AD+DE=AE=BD.

Answer 2

设BD交AC于F
∵BD⊥AE、CE⊥AE
∴BD(BF)∥CE
∴∠AFD=∠ACE
即∠AFB=∠ACE
在Rt△ABD和Rt△ACE中
AB=AC
∠AFB=∠ACE
∴Rt△ABD≌Rt△ACE
∴BD=AE，AD=CE
∴AE=AD+DE=CE+DE
∴BD=CE+DE

Answer 3

∠BAC=90°，BD⊥AE，
∴∠ABD=∠CAE,
又AB=AC,CE⊥AE,
∴△ABD≌△CAE(ASA),
∴BD=AE,AD=CE,
∴CE+DE=AD+DE=AE=BD