如图①,在三角形ABC中,∠BAC=90°,AB=AC,AE是直线,点B、C在AE的异侧,BD⊥AE于点D,CE⊥AE于点E

证明：∵∠BAD+∠ABD = 90°
∠BAD+∠CAE = 90°
∴∠ABD = ∠CAE
∵∠ADB = ∠CEA = 90°
AB = AC
∴△ABD≌△CAE
∴CE = ADBD = AE
∵AE = AD+DE
∴BD = DE+CE