Question

show that if the diagonal of a quadrilateral are equal and bisect each other at right angles then it is a square​

Answer 1

Step-by-step explanation:

let first diagonal be d¹ and let he second diagonal be d²

now let the first side be x

by Pythagoras theorem we have ,

X= √ (d¹)²+(d²)²

Now do it four every side and u will get he smae value of each side that is √(d¹)²+(d²)²

so all the sides ar equal and the figure is now square

hope it helps

Answer 2

Step-by-step explanation:

Explanation:

______________________________

Given that,

Let ABCD be a quadrilateral

It's iagonals AC and BD bisect each other at right angle at O.

To prove that

The Quadrilateral ABCD is a square.

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

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.

HenceProved!