已知点O为三角形ABC内一点,满足OA+2OB+3OC=0,求S△AOC:S△AOB:S△BOC

向量题,S△AOC:S△AOB:S△BOC=2:3:1延长OB至B',使OB'=2OB;延长OC至C',使OC'=3OC;连结B'C',取B'C'中点D,连结OD并延长至A',使DA'=OD;连结B'A',C'A',则四边形OB'A'C'为平行四边形所以2OB+3OC=OB'+OC'=OA'又因OA+2OB+3OC=0即OA+OA'=0,或AO=OA’ 所以A,O,A'三点共线,且模AO=模OA’利用同底等高三角形面积相等等得S△AOC=S△A'OC=S△OCB'=2S△BOCS△AOB=S△A'OB=S△OBC'=3S△BOC所以S△AOC:S△AOB:S△BOC=2S△BOC:3S△BOC:S△BOC=2:3:1