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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