#### Answers

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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

sorry sorry....!!dis was wrng

**Answer:** Using the same figure,

If DO=AO

- angle DAO =angle BAO=4degree

(Angles opposite to equal

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

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

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

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.

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.

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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.

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