如图,在三角形ABC中,已知角ABC与角ACB的平分线相交于点O,求证:(1)角BOC=90度+1/2角A(2)角BOC大于角A
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![如图,在三角形ABC中,已知角ABC与角ACB的平分线相交于点O,求证:(1)角BOC=90度+1/2角A(2)角BOC大于角A](/uploads/image/z/6961886-62-6.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E4%B8%AD%2C%E5%B7%B2%E7%9F%A5%E8%A7%92ABC%E4%B8%8E%E8%A7%92ACB%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E7%9B%B8%E4%BA%A4%E4%BA%8E%E7%82%B9O%2C%E6%B1%82%E8%AF%81%3A%281%29%E8%A7%92BOC%3D90%E5%BA%A6%2B1%2F2%E8%A7%92A%282%29%E8%A7%92BOC%E5%A4%A7%E4%BA%8E%E8%A7%92A)
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