题目内容
设O为△ABC的三个内角平分线的交点,当AB=AC=5,BC=6时,
=λ
+μ
(λ,μ∈R),则λ+μ=______.
| AO |
| AB |
| BC |
因为:O为△ABC内角平分线的交点,令,|AB|=c,|AC|=b,|BC|=a,则有 a×
+ b×
+c×
=0∵
=
-
,
=
-
,
=
+
,
∴a×
+ b×(
-
)+c×(
-
)=0,
∴(a+b+c)
=-(b+c)×
-c×
∴
=-
×
-
×
∵
=λ
+μ
(λ,μ∈R)
∴λ=-
,μ=-
∵a=6,b=c=5.
∴λ=-
,μ=-
∴λ+μ=-
故答案为-
.
| OA |
| OB |
| OC |
| OB |
| OA |
| BA |
| OC |
| OA |
| CA |
| AC |
| AB |
| BC |
∴a×
| OA |
| OA |
| BA |
| OA |
| CA |
∴(a+b+c)
| OA |
| AB |
| BC |
∴
| OA |
| b+c |
| a+b+c |
| AB |
| c |
| a+b+c |
| BC |
∵
| AO |
| AB |
| BC |
∴λ=-
| b+c |
| a+b+c |
| c |
| a+b+c |
∵a=6,b=c=5.
∴λ=-
| 10 |
| 16 |
| 5 |
| 16 |
∴λ+μ=-
| 15 |
| 16 |
故答案为-
| 15 |
| 16 |
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