1、证明:
∵BD⊥AE,CE⊥AE
∴∠ADB=∠AEC=90
∴∠BAD+∠ABD=90
∵∠BAC=90
∴∠BAD+∠CAE=180-∠BAC=90
∴∠ABD=∠CAE
∵AB=AC
∴△ABD≌△CAE (AAS)
∴AE=BD,AD=CE
∵AE+AD=DE
∴BD+CE=DE
2、CE+DE=BD
证明:
∵BD⊥AE,CE⊥AE
∴∠ADB=∠AEC=90
∴∠BAD+∠ABD=90
∵∠BAC=90
∴∠BAD+∠CAE=∠BAC=90
∴∠ABD=∠CAE
∵AB=AC
∴△ABD≌△CAE (AAS)
∴AE=BD,AD=CE
∵AD+DE=AE
∴CE+DE=BD
3、BD+DE=CE
证明:
∵BD⊥AE,CE⊥AE
∴∠ADB=∠AEC=90
∴∠BAD+∠ABD=90
∵∠BAC=90
∴∠BAD+∠CAE=180-∠BAC=90
∴∠ABD=∠CAE
∵AB=AC
∴△ABD≌△CAE (AAS)
∴AE=BD,AD=CE
∵AE+DE=AD
∴BD+DE=CE
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