Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given that,
OA = OC
OB = OD
and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show that,
if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
i.e., we have to prove that ABCD is parallelogram and AB = BC = CD = AD
Proof:
In ΔAOB and ΔCOB,
OA = OC (Given)
∠AOB = ∠COB (Opposite sides of a parallelogram are equal)
OB = OB (Common)
Therefore, ΔAOB ≅ ΔCOB [SAS congruency]
Thus, AB = BC [CPCT]
Similarly we can prove,
BC = CD
CD = AD
AD = AB
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.
Hence Proved.