设G是△ABC中线AD上一点,满足AG:GD=2:1D为中线,所以,OD=(OB+OC)/2OG=1/3*OA+2/3*OD=OA/3+2/3*(OB+OC)/2=(OA+OB+OC)/3同样中线BE上满足AG':G'E=2:1的点G',有OG'=(OA+OB+OC)/3中线CF上满足AG":G"F=2:1的点G",有OG"=(OA...