设D,E,F是△ABC的边BC,CA,AB的中点,则向量AD=(1/2)(AB+AC),BE=(1/2)(BC+BA),CF=(1/2)(CA+CB).设mAD+nBE=0,其中m,n是实数,由BC=BA+AC,得(m-2n)AB+(m+n)AC=0,因AB,AC不共线,故m-2n=0,m+n=0,∴m=n=0,∴AD,BE不共线,同理...