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

问题描述:

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=∠E
AB=AC

∴△ABD≌△CAE(AAS).
∴BD=AE,EC=AD.
∵AE=AD+DE,
∴BD=EC+ED.