题目内容
△ABC中,∠A=60°,∠A的平分线AD交边BC于D,已知AB=3,且
=
+λ
(λ∈R),则AD的长为( )
. |
| AD |
| 1 |
| 6 |
. |
| AB |
. |
| AC, |
A、
| ||||
B、
| ||||
| C、1 | ||||
| D、2 |
分析:D在AC上及存在实数t满足
=t
+(1-t)
可得
=
+
结合向量加法的三角形法则可得
+
=
+
从而有
=
(
-
)=
结合根据角平分线性质可得
=
可求AC
由向量的性质可得|
|=
,把已知数据代入可求
| AD |
| AB |
| AC |
| AD |
| 1 |
| 6 |
| AB |
| 5 |
| 6 |
| AC |
| AB |
| BD |
| 1 |
| 6 |
| AB |
| 5 |
| 6 |
| AC |
| BD |
| 5 |
| 6 |
| AC |
| AB |
| 5 |
| 6 |
| BC |
| AB |
| AC |
| BD |
| CD |
由向量的性质可得|
| AD |
|
解答:解:∵D在CB上
∴存在实数t满足
=t
+(1-t)
∵
=
+λ
(λ∈R),则t=
,1-t=
∵∴
=
+
∵
+
=
+
∴
=
(
-
)=
∵AD为∠A的平分线,根据角平分线性质可得
=
?
=5?AC=
∴|
|=
=
=
=
=
故选A.
∴存在实数t满足
| AD |
| AB |
| AC |
∵
. |
| AD |
| 1 |
| 6 |
. |
| AB |
. |
| AC, |
| 1 |
| 6 |
| 5 |
| 6 |
∵∴
| AD |
| 1 |
| 6 |
| AB |
| 5 |
| 6 |
| AC |
∵
| AB |
| BD |
| 1 |
| 6 |
| AB |
| 5 |
| 6 |
| AC |
∴
| BD |
| 5 |
| 6 |
| AC |
| AB |
| 5 |
| 6 |
| BC |
∵AD为∠A的平分线,根据角平分线性质可得
| AB |
| AC |
| BD |
| CD |
| 3 |
| AC |
| 3 |
| 5 |
∴|
| AD |
|
(
|
|
=
|
| ||
| 2 |
故选A.
点评:本题主要考查了平面向量的共线定理:D在AB上,存在实数t满足
=t
+(1-t)
的应用,角平分线的性质定理的综合应用,平面向量数量积的性质,本题的综合性比较强,运用的知识较为灵活.
| AD |
| AB |
| AC |
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