Answer :

Given:- Quadrilateral OACB with diagonals bisect each other at 90°.

Proof:-It is given diagonal of a quadrilateral bisect each other

Therefore, by property of parallelogram (i.e. diagonal bisect each other) this quadrilateral must be a parallelogram.

Now as Quadrilateral OACB is parallelogram, its opposite sides must be equal and parallel.

OA = BC and AC = OB

Let, O is at origin.

are position vector of A and B

Therefore from figure, by parallelogram law of vector addition

And, by triangular law of vector addition

As given diagonal bisect each other at 90°

Therefore AB and OC make 90° at their bisecting point D

Or, their dot product is zero

Hence we get

OA = AC = CB = OB

i.e. all sides are equal

Therefore by property of rhombus i.e

Diagonal bisect each other at 90°

And all sides are equal

Quadrilateral OACB is a rhombus

Hence, proved.

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