a^4+b^4+c^2=(ab)^2+(bc)^2+(ac)^2
2a^4+2b^4+2c^2-2(ab)^2-2(bc)^2-2(ac)^2=0
a^4-2(ab)^2+b^4+a^4-2(ac)^2+c^2+b^4-2(bc)^2+c^2=0
(a^2-b^2)^2+(a^2-c^2)^2+(b^2-c^2)^2=0
则a^2-b^2=0 a^2-c^2=0 b^2-c^2=0
所以a=b=c
所以是等边三角形
a^4+b^4+c^2=(ab)^2+(bc)^2+(ac)^2
2a^4+2b^4+2c^2-2(ab)^2-2(bc)^2-2(ac)^2=0
a^4-2(ab)^2+b^4+a^4-2(ac)^2+c^2+b^4-2(bc)^2+c^2=0
(a^2-b^2)^2+(a^2-c^2)^2+(b^2-c^2)^2=0
则a^2-b^2=0 a^2-c^2=0 b^2-c^2=0
所以a=b=c
所以是等边三角形