题目内容

设点O在△ABC内部,且有
OA
+2
OB
+3
OC
=
0
,则△AOB,△AOC,△BOC的面积比为(  )
A.1:2:3B.3:2:1C.2:3:4D.4:3:2
如图所示,
延长OB到点E,使得
OE
=2
OB

分别以
OA
OE
为邻边作平行四边形OAFE.
OA
+2
OB
=
OA
+
OE
=
OF

OA
+2
OB
+3
OC
=
0

OF
=3
OC

AF
=2
OB

DF
=2
OD

CO
=
OD

∴S△ABC=2S△AOB
同理可得:S△ABC=3S△AOC,S△ABC=6S△BOC
∴△AOB,△AOC,△BOC的面积比=3:2:1.
故选:B.
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