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．∵在△ABD和△CAE中∠ABD＝∠EAC∠BDA＝∠EAB＝AC，∴△ABD≌△CAE（AAS）．∴BD=AE，EC=AD．∵AE=A...