解：因为a^2+b^2c^2-ab-ac-bc=0乘以2，得（a-b)^2+(a-c)^2+(b-c)^2=0∴a-b=0 b-c=0 a-c=0∴a=b=c∴等边三角形
a^2+b^2+c^2-ab-bc-ca=0.5[(a-b)^2+(b-c)^2+(c-a)^2]=0=>(a-b)^2+(b-c)^2+(c-a)^2=0 => a=b,b=c,c=a 等边 △
