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9.D【解析】不妨建立如图所示的平面直角坐标系,OA=a, OB=b, OC=C, OD=a+b, A(2, 0), B(2, 2)D(4,2),C在直线OD上,令OM=4a,M(8,0),所以14a-c|=|OM-OC|=|CM|,即求CM的最小值,即为点M到直线OD的距离d,直线OD的方程为y2x,由点到直线的距离公式可得d=5…故选D项B(2,2)D(4,20)8,0)x
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