011-40705070  or  
Call me
Download our Mobile App
Select Board & Class
  • Select Board
  • Select Class
Kalai from KENDRIYA VIDYALAYA , asked a question
Subject: Math , asked on 5/12/12

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

pls explain.

EXPERT ANSWER

Ankush Jain ,Meritnation Expert added an answer
Answered on 5/12/12

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 ∠AOB = ∠BOC = ∠COD = ∠AOD = 90º. 

Now, to prove ABCD is a square, we have to prove that ABCD is a parallelogram, AB = BC = CD = AD, and one of its interior angles is 90º.

Proof:

In ΔAOB and ΔCOD,

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

∠AOB = ∠COD (Vertically opposite angles)

∴ ΔAOB ≅ ΔCOD (SAS congruence rule)

∴ AB = CD (By CPCT) ... (1)

And, ∠OAB = ∠OCD (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 ΔAOD and ΔCOD,

AO = CO (Diagonals bisect each other)

∠AOD = ∠COD (Given that each is 90º)

OD = OD (Common)

∴ ΔAOD ≅ ΔCOD (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 ΔADC and ΔBCD,

AD = BC (Already proved)

AC = BD (Given)

DC = CD (Common)

∴ ΔADC ≅ ΔBCD (SSS Congruence rule)

∴ ∠ADC = ∠BCD (By CPCT)

However, ∠ADC + ∠BCD = 180° (Co-interior angles)

⇒ ∠ADC + ∠ADC = 180°

⇒ 2∠ADC = 180°

⇒ ∠ADC = 90°

One of the interior angles of quadrilateral ABCD is a right angle.

Thus, we have obtained that ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angles is 90º. Therefore, ABCD is a square.

This conversation is already closed by Expert

View More
Sanika , From Chanderbala Modi Academy , added an answer
Answered on 9/10/11

I WILL SOON GVE U DIS ANS

Sanika , From Chanderbala Modi Academy , added an answer
Answered on 9/10/11

can u wait?

Sanika , From Chanderbala Modi Academy , added an answer
Answered on 9/10/11

Using the same figure,

If DO=AO

quadrilaterals 7
 

(Angles opposite to equal sides are equal)

So, all angles of the quadrilateral are right angles making it a square.

6. Diagonal AC of a parallelogram ABCD bisects angle A . Show that

(i) it bisects angle C also,

(ii) ABCD is a rhombus.

quadrilaterals 8
 

Answer: ABCD is a parallelogram where diagonal AC bisects angle DAB

quadrilaterals 9
 

As diagonals are intersecting at right angles so it is a rhombus

7. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ. Show that:

quadrilaterals 10
quadrilaterals 11
quadrilaterals 12
 

With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram

Sanika , From Chanderbala Modi Academy , added an answer
Answered on 9/10/11

sorry sorry....!!dis was wrng

Sanika , From Chanderbala Modi Academy , added an answer
Answered on 9/10/11

Answer: Using the same figure,

If DO=AO
 

  1. angle DAO =angle BAO=4degree

(Angles opposite to equal   sides are equal)

So, all angles of the quadrilateral are right angles making it a square

Sanika , From Chanderbala Modi Academy , added an answer
Answered on 9/10/11

here,u make a sqyare ABCD and make its diagnols which intersect at o.

Payal Aggarwal , From Vidya Niketan High School No.2 , added an answer
Answered on 9/10/11

 It has a simple solution.

Diagonals are equal only in sqaure and rhombus

And

Diagonals bisect each other at right angle only in square and rectangle

So, Square is the only figure in which diagonals are equal and bisect each other at right angle

Param Sukhadia , From Kendriya Vidhyalaya No. 1 , added an answer
Answered on 17/11/12

      in //gm ABCD, AC and BD are diagonals of parallelogram

PROVE:OA=OC & OB=OD

ABCD is a parallelogram ,therefore

AB//DC and AD//BC

in triangle AOB and COD

 

AB=CD(opp sides of parallelogram are equal)

 

therefore triangle AOBtriangleCOD(ASA congurence criterian)

OA=OC & OB=OD(cpct)

hence, prove that diangonals of parallelogram bisect each other.    in //gm ABCD, AC and BD are diagonals of parallelogram

PROVE:OA=OC & OB=OD

ABCD is a parallelogram ,therefore

AB//DC and AD//BC

in triangle AOB and COD

 

AB=CD(opp sides of parallelogram are equal)

 

therefore triangle AOBtriangleCOD(ASA congurence criterian)

OA=OC & OB=OD(cpct)

hence, prove that diangonals of parallelogram bisect each other.

Param Sukhadia , From Kendriya Vidhyalaya No. 1 , added an answer
Answered on 17/11/12

      in //gm ABCD, AC and BD are diagonals of parallelogram

PROVE:OA=OC & OB=OD

ABCD is a parallelogram ,therefore

AB//DC and AD//BC

in triangle AOB and COD

 

AB=CD(opp sides of parallelogram are equal)

 

therefore triangle AOBtriangleCOD(ASA congurence criterian)

OA=OC & OB=OD(cpct)

hence, prove that diangonals of parallelogram bisect each other.

U Suck , added an answer
Answered on 4/12/12

 shwetha i can feel your tits from here