已知:向量AD=1/3向量AB+2/3向则(省略向量二字):AC+CD=AD,即CD=AD-AC=[(1/3)AB+(2/3)AC]-AC==(1/3)AB-(1/3)AC=(1/3)[AB-AC]同理:CB=CA+AB=AB-AC.CD=(1&发尝篡妒诂德磋泉单沪#47;3)CB.即知B,C,D共线.