已知三角形ABC的三边为a,b,c,所对角为A,B.C,G为三角形ABC重心,且a*向量GA+根号3*b*向量GB+根号3*c*向量GC=0向量(1)求证:向量GA+向量GB+向量GC=0向量(2)求角A
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![已知三角形ABC的三边为a,b,c,所对角为A,B.C,G为三角形ABC重心,且a*向量GA+根号3*b*向量GB+根号3*c*向量GC=0向量(1)求证:向量GA+向量GB+向量GC=0向量(2)求角A](/uploads/image/z/6870441-57-1.jpg?t=%E5%B7%B2%E7%9F%A5%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E7%9A%84%E4%B8%89%E8%BE%B9%E4%B8%BAa%2Cb%2Cc%2C%E6%89%80%E5%AF%B9%E8%A7%92%E4%B8%BAA%2CB.C%2CG%E4%B8%BA%E4%B8%89%E8%A7%92%E5%BD%A2ABC%E9%87%8D%E5%BF%83%2C%E4%B8%94a%2A%E5%90%91%E9%87%8FGA%2B%E6%A0%B9%E5%8F%B73%2Ab%2A%E5%90%91%E9%87%8FGB%2B%E6%A0%B9%E5%8F%B73%2Ac%2A%E5%90%91%E9%87%8FGC%3D0%E5%90%91%E9%87%8F%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E5%90%91%E9%87%8FGA%2B%E5%90%91%E9%87%8FGB%2B%E5%90%91%E9%87%8FGC%3D0%E5%90%91%E9%87%8F%EF%BC%882%EF%BC%89%E6%B1%82%E8%A7%92A)
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