题目内容
在△ABC中,|
|=4,|
|=2,D是BC边上一点,
=
+
.
(1)求证:∠BAD=∠CAD;
(2)若|
|=
,求|
|的值.

AB |
AC |
AD |
1 |
3 |
AB |
2 |
3 |
AC |
(1)求证:∠BAD=∠CAD;
(2)若|
AD |
6 |
BC |

证明:(1)设
=
,
则
=
-
=
+
-
即
=
,
又∵|
|=4,
∴|
|=
,|
|=
•2=
,
又由ED∥AC,
可得∠BAD=∠EDA=∠CAD
(2)由|
|=
得
6=|
|2=(
+
)2=
•16+
•
+
•4⇒
•
=
,
则|
|2=(
-
)2=4-2
•
+16=9⇒
=3
AE |
1 |
3 |
AB |
则
ED |
AD |
AE |
1 |
3 |
AB |
2 |
3 |
AC |
1 |
3 |
AB |
即
ED |
2 |
3 |
AC |
又∵|
AB |
∴|
AE |
4 |
3 |
ED |
2 |
3 |
4 |
3 |
又由ED∥AC,
可得∠BAD=∠EDA=∠CAD
(2)由|
AD |
6 |
6=|
AD |
1 |
3 |
AB |
2 |
3 |
AC |
1 |
9 |
4 |
9 |
AB |
AC |
4 |
9 |
AB |
AC |
11 |
2 |
则|
BC |
AC |
AB |
AC |
AB |
BC |
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