Question

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

Answer 1

$\large\sf\underline\red{Given:-}</p><p>$

The diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.

$\large\sf\underline\green{Explanation:-} </p><p>$

Let ABCD be a quadrilateral and its diagonals AC and BD bisect each other at right angle at O.

$\large\sf\underline\pink{To\:prove:-}$

The Quadrilateral ABCD is a square.

$\large\sf\underline\purple{Proof:-}$

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite)

OB = OD (Diagonals bisect each other)

ΔAOB ≅ ΔCOD [SAS congruency]

__________________

Thus,

AB = CD [CPCT] — (i)

also,

∠OAB = ∠OCD (Alternate interior angles)

⇒ AB || CD

$\large\sf\underline\orange{Now:-}$

In ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Vertically opposite)

OD = OD (Common)

ΔAOD ≅ ΔCOD [SAS congruency]

_________________

Thus,

AD = CD [CPCT] — (ii)

Also,

AD = BC and AD = CD

⇒ AD = BC = CD = AB — (ii)

also, ∠ADC = ∠BCD [CPCT]

and ∠ADC + ∠BCD = 180° (co-interior angles)

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90° — (iii)

One of the interior angles is right angle.

Thus, from (i), (ii) and (iii) given quadrilateral ABCD is a square.

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Answer 2

Step-by-step 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   ..........(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  

∠B+∠B= 180  

 ∠B= 90  

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

 So, ABCD is a square.

Hence proved.