如图,AB=AC,∠BAC=90°,BD⊥AE于D,CE⊥AE于E,且BD>CE.求证:BD=EC+ED.

证明：∵∠BAC=90°,CE⊥AE,BD⊥AE,∴∠ABD+∠BAD=90°,∠BAD+∠DAC=90°,∠ADB=∠AEC=90°．∴∠ABD=∠DAC．又∵AB=AC,∴△ABD≌△CAE（AAS）．∴BD=AE,EC=AD．∵AE=AD+DE,∴BD=EC+ED