Question

Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a Rhombus.​

Answer 1

Step-by-step explanation:

Take quadrilateral ABCD , AC and BD are diagonals which intersect at O.

In △AOB and △AOD

DO=OB ∣ O is the midpoint

AO=AO ∣ Common side

∠AOB=∠AOD ∣ Right angle

So, △AOB≅△AOD

So, AB=AD

Similarly, AB=BC=CD=AD can be proved which means that ABCD is a rhombus.

solution

Answer 2

Answer:

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.