Question

show that if the digonals of a quadriateral bisect each other at right angles ,then it is a rhombus

Answer 1

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.
Given,
OA = OC, OB = OD and ∠AOB = ∠BOC = ∠OCD = ∠ODA = 90°
To show,
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 by SAS congruence condition.
Thus, AB = BC (by CPCT)
Similarly we can prove,
AB = BC = CD = AD
Opposites sides of a quadrilateral are equal hence ABCD is a parallelogram.
Thus, ABCD is rhombus as it is a parallelogram whose diagonals intersect at right angle.

Answer 2

Step-by-step explanation:

The diagonals are AC and BD

Let the diagonals bisect each other at O.

In ΔAOBandΔAOD

OA=OA (common)

OB=OD (given the bisect)

∠AOB=∠AOD (each 90° )

∴ΔAOB≅ΔAOD (SAS criteria)

The corresponding parts are equal.

AB=AD

Similarly, AB=BC

BC=CD

CD=AD

∴AB=BC=CD=DA

i.e. the quadrilateral is a Rhombus