△ABC中,∠ACB=90°,AC=BC,点E在BC上,过点C作CF⊥AE于F,延长CF使CD=AE,连接BD.求证:BD⊥BC

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△ABC中,∠ACB=90°,AC=BC,点E在BC上,过点C作CF⊥AE于F,延长CF使CD=AE,连接BD.求证:BD⊥BC
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△ABC中,∠ACB=90°,AC=BC,点E在BC上,过点C作CF⊥AE于F,延长CF使CD=AE,连接BD.求证:BD⊥BC
△ABC中,∠ACB=90°,AC=BC,点E在BC上,过点C作CF⊥AE于F,延长CF使CD=AE,连接BD.求证:BD⊥BC

△ABC中,∠ACB=90°,AC=BC,点E在BC上,过点C作CF⊥AE于F,延长CF使CD=AE,连接BD.求证:BD⊥BC
AC=BC
AE=CD
∠DCB+∠AEC=90°
∠CAE+∠AEC=90°得∠DCB=∠CAE
即由边角边得三角形ACE全等于三角形CDB
∠ACB=90° 所以∠CBD=90°所以BD⊥BC