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

PQRS is a parallelogram, let PS = A , SR = B ,  QS = D  ,SQ = C  (all bold letters are vectors)

applying triangle of vector addition  in triangle in triangle  PSR 

 B+C = A

or,  C = A - B

similarly  in triangle SQR

 B+QR = D

 B+A = D  (since QR = A )

now it is given that C is perpendicular to D

 C.D = 0  

or,  (A-B).(A+B) = 0

or  A = B

or  PS = SR 

Similarly  RQ = PQ

so  PQ = QR = RS = SP

this implies that parallelogram PQRS is a Rhombus