作业帮三角形ABC的三角平分线AA`+BB`+CC`=0向量 证明三角形是正三角形
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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
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
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
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
AA"=1/2(AB+AC) BB"=(BA+BC)1/2 CC"=(CB+CA)*1/2