△ABC的三边满足a4+b2c2-a2c2-b4=0,请判别△ABC的形状.

问题描述:

△ABC的三边满足a4+b2c2-a2c2-b4=0,请判别△ABC的形状.

a4+b2c2-a2c2-b4
=(a4-b4)+(b2c2-a2c2
=(a2+b2)(a2-b2)-c2(a2-b2
=(a2-b2)(a2+b2-c2
=(a+b)(a-b)(a2+b2-c2)=0,
∵a+b>0,
∴a-b=0或a2+b2-c2=0,
即a=b或a2+b2=c2
则△ABC为等腰三角形或直角三角形.