Question

Let ABCD be a parallelogram whose diagonals intersect at Pand let O be the origin, then OA+Ob+OC+OD equals​

Answer 1

Answer:

4 OP

Step-by-step explanation:

P which is the intersection of diagonals of parallelogram its bisects the diagonal

Thus

OP=

2

(OA+OC)

​

i.e.

OA+OC=2OP    ----------------------  (i)

Similarly

OB+OD=OP    -------------------   (ii)

Adding (i) and (ii)

we get

OA + Ob+ OC + OD

=4 OP

hope it helps you...