题目内容
设点O在△ABC内部,且有
+2
+3
=
,则△AOB,△AOC,△BOC的面积比为( )
OA |
OB |
OC |
0 |
A.1:2:3 | B.3:2:1 | C.2:3:4 | D.4:3:2 |
如图所示,
延长OB到点E,使得
=2
,
分别以
,
为邻边作平行四边形OAFE.
则
+2
=
+
=
,
∵
+2
+3
=
,
∴
=3
.
又
=2
,
∴
=2
.
∴
=
,
∴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.
延长OB到点E,使得
OE |
OB |
分别以
OA |
OE |
则
OA |
OB |
OA |
OE |
OF |
∵
OA |
OB |
OC |
0 |
∴
OF |
OC |
又
AF |
OB |
∴
DF |
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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