Question

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

Answer 1

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.