Question

The diagonals of a quadrilateral are equal and bisect each other , then the quadrilateral is :​

Answer 1

Answer:

Step-by-step explanation:

Let a Quadrilateral, ABCD whose diagonals bisect at O.

Given,

Diagonals are equal

AC=BD                  (1)

and the diagonals bisect each other at right angles

OA=OC;OB=OD           (2)

∠AOB= ∠BOC= ∠COD=  ∠AOD= 90 degrees    (3)

 Proof:

Consider △AOB and △COB

OA=OC   [from (2)]

∠AOB= ∠COB

OB is the common side

Therefore,

△AOB≅ △COB

From SAS criteria, AB=CB

Similarly, we prove

△AOB≅ △DOA, so AB=AD

△BOC≅ △COD, so CB=DC

So, AB=AD=CB=DC              (4)

So, in quadrilateral ABCD, both pairs of opposite sides are equal, hence ABCD is  parallelogram

In △ABC and △DCB

AC=BD            (from (1))

AB=DC            (from (4))

BC is the common side

△ABC≅ △DCB

So, from SSS criteria, ∠ABC= ∠DCB

Now,

AB∥CD

∠B+∠C= 180 degrees  

 

∠B+∠B= 180 degrees  

 

∠B= 90 degrees  

 

Hence, ABCD is a parallelogram with all sides equal and one angle is 90degrees  

 

So, ABCD is a square.

Hence proved

Answer 2

Answer:

Step-by-step explanation:

Let a Quadrilateral, ABCD whose diagonals bisect at O.

Given,

Diagonals are equal

AC=BD                  (1)

and the diagonals bisect each other at right angles

OA=OC;OB=OD           (2)

∠AOB= ∠BOC= ∠COD=  ∠AOD= 90 degrees    (3)

Proof:

Consider △AOB and △COB

OA=OC   [from (2)]

∠AOB= ∠COB

OB is the common side

Therefore,

△AOB≅ △COB

From SAS criteria, AB=CB

Similarly, we prove

△AOB≅ △DOA, so AB=AD

△BOC≅ △COD, so CB=DC