a(1-2cosA)+b(1-2cosB)+c(1-2cosC)=0
求证ABC为正三角形

cosA=(b^2+c^2-a^2)/(2bc)
cosB=(c^2+a^2-b^2)/(2ca)
cosC=(a^2+b^2-c^2)/(2ab)

a(1-2cosA)+b(1-2cosB)+c(1-2cosC)
=a[1-(b^2+c^2-a^2)/(bc)]+b[1-(c^2+a^2-b^2)/ca)]+c[1-(a^2+b^2-c^2)/(ab)]=0

*abc

a^2bc-a^2b^2-a^2c^2+a^4+b^2ac-b^2c^2-a^2b^2+b^4+c^2ab-c^2a^2-c^2b^2+c^4=0

*2

(a^4+b^4-2a^2b^2)+(a^4+c^4-2a^2c^2)+(b^4+c^4-2b^2c^2)-(a^2b^2+a^2c^2-2a^2bc)-(a^2^b^2+b^2c^2-2b^2ac)-(c^2a^2+c^2b^2-2c^2ab)=0

(a^2-b^2)^2+(a^2-c^2)^2+(b^2-c^2)^2-(ab-ac)^2-(ab-bc)^2-(ca-bc)^2=0

(a+b)^2(a-b)^2+(a+c)^2(a-c)^2+(b+c)^2(b-c)^2-a^2(b-c)^2-b^2(a-c)^2-c^2(a-b)^2=0

(a-b)^2[(a+b)^2-c^2]+(a-c)^2[(a+c)^2-b^2]+(b-c)^2[(b+c)^2-a^2]=0

a+b>c,a+c>b,b+c>a
上式三个乘积均>=0

所以:
a-b=0.a-c=0,b-c=0
a=b=c

证明角相等也可:
a=2RsinA
b=2RsinB
c=2RsinC
a(1-2cosA)+b(1-2cosB)+c(1-2cosC)
=2RsinA(1-2cosA)+2RsinB(1-2cosB)+2RsinC(1-2cosC)=0

2sinA-2sin2A+2sinB-2sin2A+2sin2B-2sin2C=0

2sinA+2sinB+2sinC-(sin2A+sin2B)-(sin2B+sin2C)-(sin2A+sin2B)=0

2sinA+2sinB+2sinC-2sin(A+B)cos(A-B)-2sin(B+C)cos(B-C)-2sin(C+A)cos(C-A)=0

2sinA+2sinB+2sinC-2sinCcos(A-B)-2sinAcos(B-C)-2sinBcos(C-A)=0

sinA[1-cos(B-C)]+sinB[1-cos(C-A)]+sinC[1-cos(A-B)]=0

sinA>0,sinB>0,sinC>0,
1-cos(B-C)>=0,1-cos(C-A)>=0,1-cos(A-B)>=0

cos(B-C)=1,B=C
cos(C-A)=1,C=A
cos(A-B)=1,A=B

A=B=C