已知,如图,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,使AD=AB,连接DC.求证;DC⊥BC

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已知,如图,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,使AD=AB,连接DC.求证;DC⊥BC

已知,如图,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,使AD=AB,连接DC.求证;DC⊥BC
已知,如图,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,
在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,使AD=AB,连接DC.求证;DC⊥BC

已知,如图,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,在△ABC中,AB=AC,AE⊥BC于点E,延长BA至D,使AD=AB,连接DC.求证;DC⊥BC
证明:
∵AB=AC
∴∠B=∠BCA
又∠BAC+∠B+∠BCA=180°
∴∠BAC=180°-2∠BCA
又AD=AC=AB
∴∠D=∠ACD
又∠BAC =∠D+∠ACD
∴∠BAC=2∠ACD
∴180°-2∠BCA=2∠ACD
180°=2(∠BCA+∠ACD)
∠BCA+∠ACD=90°
∴DC⊥BC