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Prove that if the diagonals of a quadrilateral bisect each other at right angles then it is a rhombus.

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#### Solution

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle.

i.e. OA = OC, OB = OD

And, ∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.

Now, in ΔAOD and DCOD

OA = OC (Diagonal bisects each other)

∠AOD = ∠COD ...(Each 90°)

OD = OD ...(common)

∴ΔAOD ≅ΔCOD ...(By SAS congruence rule)

∴ AD = CD ….(i)

Similarly, we can prove that

AD = AB and CD = BC ….(ii)

From equations (i) and (ii), we can say that

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, so, we can say that ABCD is a parallelogram.

Since all sides of a parallelogram ABCD are equal, so, we can say that ABCD is a rhombus.

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