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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