a^2+b^2+c^2=ab+bc+ac
a^2+b^2+c^2-ab-bc-ac=0
2a^2+2b^2+2c^2-2ab-2bc-2ac=0
(a^2-2ab+b^2)+(a^2-2ac+c^2)+(b^2-2bc+c^2)=0
(a-b)^2+(a-c)^2+(b-c)^2=0
因为(a-b)^2>=0,(a-c)^2>=0,(b-c)^2>=0,
要使(a-b)^2+(a-c)^2+(b-c)^2=0成立,则(a-b)^2=0且(a-c)^2 =0且(b-c)^2=0,当且仅当三者取零时才满足!
解之得a=b,a=c,b=c
即：a=b=c
故三角形ABC为等边三角形
希望能帮到你哈!