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. Posted by Ankush Jain(MeritNation Expert)on 5/12/12 This conversation is already closed by Expert

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. Posted by Ankush Jain(MeritNation Expert)on 5/12/12 This conversation is already closed by Expert

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. Posted by Ankush Jain(MeritNation Expert)on 5/12/12 This conversation is already closed by Expert

Using the same figure, If DO=AO (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. Answer: ABCD is a parallelogram where diagonal AC bisects angle DAB 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: With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram Posted by Sanika(student)on 9/10/11

Using the same figure, If DO=AO (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. Answer: ABCD is a parallelogram where diagonal AC bisects angle DAB 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: With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram Posted by Sanika(student)on 9/10/11

Using the same figure, If DO=AO (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. Answer: ABCD is a parallelogram where diagonal AC bisects angle DAB 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: With equal opposite angles and equal opposite sides it is proved that APCQ is a parallelogram Posted by Sanika(student)on 9/10/11

Answer: Using the same figure, If DO=AO 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 Posted by Sanika(student)on 9/10/11

Answer: Using the same figure, If DO=AO 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 Posted by Sanika(student)on 9/10/11

Answer: Using the same figure, If DO=AO 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 Posted by Sanika(student)on 9/10/11

here,u make a sqyare ABCD and make its diagnols which intersect at o. Posted by Sanika(student)on 9/10/11

here,u make a sqyare ABCD and make its diagnols which intersect at o. Posted by Sanika(student)on 9/10/11

here,u make a sqyare ABCD and make its diagnols which intersect at o. Posted by Sanika(student)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 Posted by Payal Aggarwal(student)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 Posted by Payal Aggarwal(student)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 Posted by Payal Aggarwal(student)on 9/10/11

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 AOB= triangleCOD(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 AOB= triangleCOD(ASA congurence criterian) OA=OC & OB=OD(cpct) hence, prove that diangonals of parallelogram bisect each other. Posted by Param Sukhadia(student)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 AOB= triangleCOD(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 AOB= triangleCOD(ASA congurence criterian) OA=OC & OB=OD(cpct) hence, prove that diangonals of parallelogram bisect each other. Posted by Param Sukhadia(student)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 AOB= triangleCOD(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 AOB= triangleCOD(ASA congurence criterian) OA=OC & OB=OD(cpct) hence, prove that diangonals of parallelogram bisect each other. Posted by Param Sukhadia(student)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 AOB= triangleCOD(ASA congurence criterian) OA=OC & OB=OD(cpct) hence, prove that diangonals of parallelogram bisect each other. Posted by Param Sukhadia(student)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 AOB= triangleCOD(ASA congurence criterian) OA=OC & OB=OD(cpct) hence, prove that diangonals of parallelogram bisect each other. Posted by Param Sukhadia(student)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 AOB= triangleCOD(ASA congurence criterian) OA=OC & OB=OD(cpct) hence, prove that diangonals of parallelogram bisect each other. Posted by Param Sukhadia(student)on 17/11/12