Question

show that if the diagonal of a quadrilateral bisect each other at right angle,then it is a rhombus​

Answer 1

Step-by-step explanation:

We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ΔAOB and ΔAOD, we have

AO = AO

[Common]

OB = OD

[Given that O in the mid-point of BD]

∠AOB = ∠AOD

[Each = 90°]

ΔAOB ≌ ΔAOD

[SAS criteria]

Their corresponding parts are equal.

AB = AD...(1)Similarly,AB = BC...(2) BC = CD...(3) CD = AD...(4)

∴ From (1), (2), (3) and (4), we have AB = BC CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Answer 2

Answer:-

Let ABCD be a quadrilateral whose diagonals bisect each other at right angles.

★ Given that:-

OA = OC

OB = OD

∠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.

★ 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.