By vector method, prove that if the diagonals of a parallelogram intersect perpendicularly,then the parallelogram is rombus.

Explanation

Assuming the sides of a paralleogram as `vec(OP)=vec(A)`, and `vec(OR)=vec(B)`,
its diagonals can be given as `vec(OQ)=vec(A)+vec(B)` and `vec(RP)=vec(A)-vec(B)`,
respectively.

Since `vec(OQ)_|_vec(RP),vec(OQ).vec(RP)=0,` substituting `OQ=vec(A)+vec(B)`and
`RP =vec(A)-vec(B)`,we have `(vec(A)+vec(B)).(vec(A)-vec(B))=0`.
This gives `vec(A).vec(A)-vec(A).vec(B)+vec(B).vec(A)-vec(B).vec(B)=0`
since `vec(A).vec(A)-|vec(A)|^(2),vec(B)+vec(B)=|vec(B)|^(2)`, and `vec(A).vec(B)=vec(B).vec(A)`, we have
`|vec(A)|^(2)-|vec(B)|^(2)0.` Hence , `|vec(A)|=|vec(B)|.` It means that OP=OR.
This tells us that the given parallelogram is a rombus.