设a(x1,y1),b(x2,y2),a+b=(x1+x2,y1+y2)
a-2b=(x1-2x2,y1-2y2)
根号x1^2+y1^2=2(x2^2+y2^2)
(x1+x2)(x1-2x2)+(y1+y2)(y1-2y2)=0即：
x1^2-x1x2-2x2^2+y1^2-y2y2-2y2^2=0
联立以上两式可得：
x1x2+y1y2=0,所以a与b垂直!