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.
\(\vec a\) and \(\vec b\) are position vector of A and B
Therefore from figure, by parallelogram law of vector addition
\(\vec{OC}=\vec a+\vec b\)
And, by triangular law of vector addition
\(\vec{AB}=\vec a-\vec b\)
As given diagonal bisect each other at 90°
Therefore AB and OC make 90° at their bisecting point D
⇒ ∠ ADC = ∠ CDB = ∠ BDO = ∠ ODA = 90°
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.
