题目内容

如图,O,A,B三点不共线,且
OC
=2
OA
OD
=3
OB
,设
OA
=a
OB
=b

(1)试用a,b表示向量
OE

(2)设线段AB,OE,CD的中点分别为L,M,N,试证明L,M,N三点共线.
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(1)∵B,E,C三点共线,
OE
=x
OC
+(1-x)
OB
=2x
a
+(1-x)
b
,①
同理,∵A,E,D三点共线,可得
OE
=y
a
+3(1-y)
b
,②
比较①,②,得
2x=y
1-x=3(1-y)
解得x=
2
5
,y=
4
5

OE
=
4
5
a
+
3
5
b

(2)∵
OL
=
a
+
b
2
OM
=
1
2
OE
=
4
a
+3
b
10
ON
=
1
2
(
OC
+
OD
)=
2
a
+3
b
2

MN
=
ON
-
OM
=
6
a
+12
b
10
ML
=
OL
-
OM
=
a
+2
b
10

MN
=6
ML
,∴L,M,N三点共线.
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