Question -
Answer -
Let ABCD be a quadrilateral whose diagonals bisect each other atright 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 rightangles, 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 areequal)
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 aparallelogram.
, ABCD is rhombus as it is a parallelogram whose diagonalsintersect at right angle.
HenceProved.