Question

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

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Answer 1

Explanation:

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

0

..........(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,BC is the tansversal

∠B+∠C= 180

0

∠B+∠B= 180

0

∠B= 90

0

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

0

So, ABCD is a square.

Hence proved.

Answer 2

Explanation:

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

0

..........(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,BC is the tansversal

∠B+∠C= 180

0

∠B+∠B= 180

0

∠B= 90

0

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

0

So, ABCD is a square.

Hence proved.