#abs(A+B) = sqrt(abs(A)^2+abs(B)^2+2abs(A)abs(B)costheta)# where #theta# is the angle between them. And,
#abs(A-B) = sqrt(abs(A)^2+abs(B)^2-2abs(A)abs(B)costheta)#
Now, the equality #abs(A+B)=abs(A-B)# gives:
#sqrt(abs(A)^2+abs(B)^2+2abs(A)abs(B)costheta)=sqrt(abs(A)^2+abs(B)^2-2abs(A)abs(B)costheta)#
squaring both sides, we get:
#cancel(abs(A)^2)+cancel(abs(B)^2)+2abs(A)abs(B)costheta=cancel(abs(A)^2)+cancel(abs(B)^2)-2abs(A)abs(B)costheta#
#:. cancel2abs(A)abs(B)costheta=-cancel2abs(A)abs(B)costheta#
Now if either #abs(A)# or #abs(B)# is zero or both are zero, the statement trivially follows for all #theta#.
The nontrivial case is #abs(A)!=0# and #abs(B)!=0#, then we can safely cancel them out:
#:. cancelabs(A)cancelabs(B)costheta=-cancelabs(A)cancelabs(B)costheta#
to have: #2costheta =0# and hence, #costheta =0# which is true for #theta = 90°# or #pi/2 rad#.
