三道关于相似三角形的解答题!急!1.已知,如图,三角形ABC中,AB=AC,AD是中线,P是AD上一点,过C作CF//AB,延长BP交AC于E,交CF于F,求证:BP的平方=PE*PF2.已知,如图,角C=90度,以BC为边向外作正方形BEDC,连接AE交BC
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![三道关于相似三角形的解答题!急!1.已知,如图,三角形ABC中,AB=AC,AD是中线,P是AD上一点,过C作CF//AB,延长BP交AC于E,交CF于F,求证:BP的平方=PE*PF2.已知,如图,角C=90度,以BC为边向外作正方形BEDC,连接AE交BC](/uploads/image/z/10153268-44-8.jpg?t=%E4%B8%89%E9%81%93%E5%85%B3%E4%BA%8E%E7%9B%B8%E4%BC%BC%E4%B8%89%E8%A7%92%E5%BD%A2%E7%9A%84%E8%A7%A3%E7%AD%94%E9%A2%98%21%E6%80%A5%211.%E5%B7%B2%E7%9F%A5%2C%E5%A6%82%E5%9B%BE%2C%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2CAB%3DAC%2CAD%E6%98%AF%E4%B8%AD%E7%BA%BF%2CP%E6%98%AFAD%E4%B8%8A%E4%B8%80%E7%82%B9%2C%E8%BF%87C%E4%BD%9CCF%2F%2FAB%2C%E5%BB%B6%E9%95%BFBP%E4%BA%A4AC%E4%BA%8EE%2C%E4%BA%A4CF%E4%BA%8EF%2C%E6%B1%82%E8%AF%81%3ABP%E7%9A%84%E5%B9%B3%E6%96%B9%3DPE%2APF2.%E5%B7%B2%E7%9F%A5%2C%E5%A6%82%E5%9B%BE%2C%E8%A7%92C%3D90%E5%BA%A6%2C%E4%BB%A5BC%E4%B8%BA%E8%BE%B9%E5%90%91%E5%A4%96%E4%BD%9C%E6%AD%A3%E6%96%B9%E5%BD%A2BEDC%2C%E8%BF%9E%E6%8E%A5AE%E4%BA%A4BC)
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