已知平行四边形ABCD的两条对角线AC,BD交于E.O是任意一点,求证:OA+OB+OC+OD=4OE.(OA,OB,OC,OD,OE为向量)
问题描述:
已知平行四边形ABCD的两条对角线AC,BD交于E.O是任意一点,求证:OA+OB+OC+OD=4OE.(OA,OB,OC,OD,OE为向量)
答
这个简单
OA-OE=EA
OB-OE=EB
OC-OE=EC
OD-OE=ED
OA-OE+OB-OE+OC-OE+OD-OE=EA+EB+EC+ED=0
即OA+OB+OC+OD=4OE
答
OA-OE=EA
OB-OE=EB
OC-OE=EC
OD-OE=ED
OA-OE+OB-OE+OC-OE+OD-OE=EA+EB+EC+ED=0
即OA+OB+OC+OD=4OE