Question

Show that if the diagonal of a quadrilateral are equal and bisect each other at right angle then it is a square

Answer 1

Let us consider a quadrilateral ABCD in

which the diagonals AC and BD

intersect each other at O. It is given

that the diagonals of ABCD are equal

and bisect each other at right angles.

Therefore, AC = BD, OA = OC, OB = OD,

and JACOB = XBOX = COD = LOAD =

90°

Now, to prove ABCD is a square, we

have to prove that ABCD is

parallelogram, AB = BC = CD = AD, and

one of its interior angles is 90°.

Proof:

In AAOB and ACID,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

angle AOB = angle COD(Vertically opposite

angles)

∆OBA~=∆COD (SAS congruence

rule)

AB = CD (By CPCT) (1)

And, ZOAB = ZOCD (By CPCT)

However, these are alternate interior

angles for line AB and CD and alternate

interior angles are equal to each other

only when the two lines are parallel.

AB CD... (2)

From equations (1) and (2), we obtain

ABCD is a parallelogram.

In GOD and ACID,

AO = CO (Diagonals bisect each other)

ZAID = ZAID (Given that each is 90°)

OD = OD (Common)

AAD = ACOD (SAS congruence

rule)

AD = DC (3)

However, AD = BC and AB = CD

(Opposite sides of parallelogram

ABCD)

AB = BC = CD = DA

Therefore, all the sides of quadrilateral

ABCD are equal to each other.

In AADC and ABCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

AADC = ABCD (SSS Congruence

rule)

ZADCO = ABCD (By CPCT)

However, ZADCO + ABCD = 180°

(Co-interior angles)

ZAC + ZAC = 180°

However, ZADCO + ABCD = 180°

(Co-interior angles)

LADC + ZADC = 180°

2 ADC = 180°

LADC = 90°

One of the interior angles of

quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is

parallelogram, AB = BC = CD = AD and

one of its interior angles is 90°.

Therefore, ABCD is a square.

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!