如图,在正三角形ABC中,向量AD=三分之一向量AB,向量BE=三分之一向量BC,且AE,CD交于点P,求证:BP垂直D

过A点做AF平行BC,CD交AF与F
得FD/DC = AD/DB = AF/BC = 1/2, 则FD= 1/3FC, AF = 3/4 EC
FP/PC = AF/EC = 3/4,则 PC= 4/7FC
有DP = 1/6 PC
(下列运算中为方便起见,直接将向量AD简写为ad,请注意)
bp · dc = (bd + dp) · dc = (2/3 ba + 1/7 dc) · dc = (2/3 ba + 1/7 bc - 1/7*2/3 ba) · dc
= (1/7 bc + 4/7 ba)·(bc - 2/3 ba)
= (1/7 +4/7*1/2 - 1/7 * 2/3 *1/2 - 4/7 * 2/3) * |BA| * |BA| = 0 得证

建系，不妨设三角形边长为2