在△ABC中,AB = AC,BD⊥AC于D求证:∠BAC = 2∠DBC
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![在△ABC中,AB = AC,BD⊥AC于D求证:∠BAC = 2∠DBC](/uploads/image/z/11706095-47-5.jpg?t=%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2CAB+%3D+AC%2CBD%E2%8A%A5AC%E4%BA%8ED%E6%B1%82%E8%AF%81%EF%BC%9A%E2%88%A0BAC+%3D+2%E2%88%A0DBC)
在△ABC中,AB = AC,BD⊥AC于D求证:∠BAC = 2∠DBC
在△ABC中,AB = AC,BD⊥AC于D求证:∠BAC = 2∠DBC
在△ABC中,AB = AC,BD⊥AC于D求证:∠BAC = 2∠DBC
证明:∵AB = AC,∴∠ABC =∠C,∵BD⊥交流,∴RT⊿DBC,有:∠DBC = 90?稀释眺,⊿ABC,∠BAC = 180?/ B>∠ABC,∠C = 180?∠C = 2(90?稀条),∴∠BAC = 2∠DBC.