三角形ABC的三角平分线AA`+BB`+CC`=0向量 证明三角形是正三角形
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/25 15:01:45
三角形ABC的三角平分线AA`+BB`+CC`=0向量 证明三角形是正三角形
![](http://img.wesiedu.com/upload/3/37/337b52bbf88047c41b8ff0e0608cd206.jpg)
![](http://img.wesiedu.com/upload/3/37/337b52bbf88047c41b8ff0e0608cd206.jpg)
AA'+BB'+CC' =0
let
|CA'|/|A'B| = |CA|/|AB| =k1
AA'= (AC+k1AB)/(1+k1) (1)
let
|AB'|/|B'C| = |AB|/|BC| =k2
BB'= (BA+k2BC)/(1+k2) (2)
let
|BC'|/|C'A| = |BC|/|CA| =k3
CC' = (CB+k3CA)/(1+k3) (3)
(1)+(2)+(3)
AA'+BB'+CC'= (AC+k1AB)/(1+k1) + (BA+k2BC)/(1+k2)+ (CB+k3CA)/(1+k3)
[k1/(1+k1)-1/(1+k2)]AB+ [ k2/(1+k2) - 1/(1+k3)]BC + [k3/(1+k3) -1/(1+k1)]CA =0
[1-1/(1+k1)-1/(1+k2)]AB+ [ 1-1/(1+k2) - 1/(1+k3)]BC + [1-/(1+k3) -1/(1+k1)]CA =0
cosider
AB+BC+CA =0
[1-1/(1+k1)-1/(1+k2)]:[ 1-1/(1+k2) - 1/(1+k3)]:[1-/(1+k3) -1/(1+k1)] =1:1:1
=> 1-1/(1+k1)-1/(1+k2)=1-1/(1+k2)-1/(1+k3)=1-1/(1+k3)-1/(1+k1)
=> -1/(1+k1)-1/(1+k2)=-1/(1+k2)-1/(1+k3)=-1/(1+k3)-1/(1+k1)
ie k1=k2=k3
|CA|/|AB| =k1 and |AB|/|BC| =k2 and |BC|/|CA| =k3
=> |AB|=|BC|=|CA|
=> ABC是正三角形
let
|CA'|/|A'B| = |CA|/|AB| =k1
AA'= (AC+k1AB)/(1+k1) (1)
let
|AB'|/|B'C| = |AB|/|BC| =k2
BB'= (BA+k2BC)/(1+k2) (2)
let
|BC'|/|C'A| = |BC|/|CA| =k3
CC' = (CB+k3CA)/(1+k3) (3)
(1)+(2)+(3)
AA'+BB'+CC'= (AC+k1AB)/(1+k1) + (BA+k2BC)/(1+k2)+ (CB+k3CA)/(1+k3)
[k1/(1+k1)-1/(1+k2)]AB+ [ k2/(1+k2) - 1/(1+k3)]BC + [k3/(1+k3) -1/(1+k1)]CA =0
[1-1/(1+k1)-1/(1+k2)]AB+ [ 1-1/(1+k2) - 1/(1+k3)]BC + [1-/(1+k3) -1/(1+k1)]CA =0
cosider
AB+BC+CA =0
[1-1/(1+k1)-1/(1+k2)]:[ 1-1/(1+k2) - 1/(1+k3)]:[1-/(1+k3) -1/(1+k1)] =1:1:1
=> 1-1/(1+k1)-1/(1+k2)=1-1/(1+k2)-1/(1+k3)=1-1/(1+k3)-1/(1+k1)
=> -1/(1+k1)-1/(1+k2)=-1/(1+k2)-1/(1+k3)=-1/(1+k3)-1/(1+k1)
ie k1=k2=k3
|CA|/|AB| =k1 and |AB|/|BC| =k2 and |BC|/|CA| =k3
=> |AB|=|BC|=|CA|
=> ABC是正三角形
已知abc是三角形的三边,且aa+bb+cc=ab+bc+ac,求三角形的形状
abc为三角形三边 aa+bb+cc-cb-ab-ac=0 是什么三角形
已知a.b.c是三角形的三边求(aa+bb-cc)(aa+bb-cc)-4aabb
已知a.b.c为三角形ABC的三条边,且满足aa+bb+cc-ab-bc-bc=0试判断三角形的形状
在三角形ABC中,已知aa-cc+bc=bb,则A=?
已知abc为三角形ABC的三个内角的对应边,试证明(aA+bB+cC)/(a+b+c)<π/2
若三角形的三边ABC,满足条件AA+BB+CC+338=10A+24B+26C,试判断三角形的ABC的形状.
三角形ABC的三边a,b,c满足aa+bb+cc+338=10a+24b+26c,求三角形ABC的面积
若a,b,c为三角形ABC的三边,且满足aa+bb+cc=ab+ac+bc,试判断三角形ABC的形状
I是三角形ABC的内心,射线AI、BI、CI交三角形的外接圆于A’、B’、C’.求证:AA’+BB’+CC’大于BC+C
abc为三角形3边,ABC为所对的角,求(aA+bB+cC)/(a+b+c)的范围
已知a.b.c是三角形的三边,且满足(a+b+c)(a+b+c)=3(aa+bb+cc),求证;这个三角形是等腰三角形