# Class 9 NCERT Solutions Maths Chapter 8 - Quadrilaterals

Learn to provide Mathematical proofs easily with TopperLearning’s NCERT Solutions for CBSE Class 9 Mathematics Chapter 8 Quadrilaterals. Can a parallelogram be a rhombus? How? Learn by practising with our solutions by Maths experts. Understand how to express your understanding of quadrilaterals, such as a parallelogram and rhombus, and provide clear explanations in your answers.

You can score full marks in questions based on quadrilaterals by revising key concepts with our CBSE Class 9 Maths textbook solutions. If you find any of the concepts to be mind-boggling, discuss your doubts with Maths experts through our ‘UnDoubt’ platform.

Ex. 8.1

Ex. 8.2

## Quadrilaterals Exercise Ex. 8.1

### Solution 1

Let the common ratio between the angles is x. So, the angles will be 3x, 5x, 9x and 13x respectively.

Since the sum of all interior angles of a quadrilateral is 360.

3x + 5x + 9x + 13x = 360

Since the sum of all interior angles of a quadrilateral is 360.

3x + 5x + 9x + 13x = 360

30x = 360

x = 12

Hence, the angles are

3x = 3 12 = 36

5x = 5 12 = 60

9x = 9 12 = 108

13x = 13 12 = 156

^{o}

### Solution 2

Let ABCD be a parallelogram. To show ABCD a rectangle, only we need to prove one of its interior angle is 90.

In ABC and DCB

AB = DC (opposite sides of a parallelogram are equal)

BC = BC (common)

AC = DB (given)

ABC DCB (by SSS Congruence rule)

In ABC and DCB

AB = DC (opposite sides of a parallelogram are equal)

BC = BC (common)

AC = DB (given)

ABC DCB (by SSS Congruence rule)

ABC = DCB

We know that sum of measures of angles on the same side of transversal is 180º.

ABC + DCB = 180 (AB || CD)

ABC + ABC = 180

ABC = 180

ABC = 90

Since ABCD is a parallelogram and one of its interior angles is 90, therefore, ABCD is rectangle.

### Solution 3

Let ABCD be a quadrilateral, whose diagonals AC and BD bisect each other at right angle

i.e. OA = OC, OB = OD and AOB = BOC = COD = AOD = 90

To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.

Now, in AOD and COD

OA = OC (Diagonal bisects each other)

AOD = COD (given)

OD = OD (common)

i.e. OA = OC, OB = OD and AOB = BOC = COD = AOD = 90

To prove ABCD a rhombus, we need to prove ABCD is a parallelogram and all sides of ABCD are equal.

Now, in AOD and COD

OA = OC (Diagonal bisects each other)

AOD = COD (given)

OD = OD (common)

AOD COD (by SAS congruence rule)

AD = CD (1)

Similarly we can prove that

AD = AB and CD = BC (2)

From equations (1) and (2), we can say that

AB = BC = CD = AD

Since opposite sides of quadrilateral ABCD are equal, so, we can say that ABCD is a parallelogram. Since all sides of a parallelogram ABCD are equal, so, we can say that ABCD is a rhombus.

### Solution 4

Let ABCD be a square. Let the diagonals AC and BD intersect each other at a point O.

To show diagonals of a square are equal and bisect each other at right angles, we need to prove AC = BD, OA = OC, OB = OD and AOB = 90

Now, in ABC and DCB

AB = DC (sides of square are equal to each other)

ABC = DCB (all interior angles are of 90 )

BC = BC (common side)

ABCDCB (by SAS congruency)

AC = DB (by CPCT)

To show diagonals of a square are equal and bisect each other at right angles, we need to prove AC = BD, OA = OC, OB = OD and AOB = 90

Now, in ABC and DCB

AB = DC (sides of square are equal to each other)

ABC = DCB (all interior angles are of 90 )

BC = BC (common side)

ABCDCB (by SAS congruency)

AC = DB (by CPCT)

Hence, the diagonals of a square are equal in length

Now in AOB and COD

AOB = COD (vertically opposite angles)

ABO = CDO (alternate interior angles)

AB = CD (sides of square are always equal)

AOB COD (by AAS congruence rule)

AO = CO and OB = OD (by CPCT)

Now in AOB and COD

AOB = COD (vertically opposite angles)

ABO = CDO (alternate interior angles)

AB = CD (sides of square are always equal)

AOB COD (by AAS congruence rule)

AO = CO and OB = OD (by CPCT)

Hence, the diagonals of a square bisect each other

Now in AOB and COB

Now as we had proved that diagonals bisect each other

So, AO = CO

AB = CB (sides of square are equal)

BO = BO (common)

AOB COB (by SSS congruence)

AOB = COB (by CPCT)

But, AOB + COB = 180 (linear pair)

2AOB = 180

AOB = 90

Hence, the diagonals of a square bisect each other at right angle.

### Solution 5

Let us consider a quadrilateral ABCD in which the diagonals AC and BD intersect each other at O.

Given that the diagonals of ABCD are equal and bisect each other at right angles.

Given that the diagonals of ABCD are equal and bisect each other at right angles.

So, AC = BD, OA = OC, OB = OD and

AOB = BOC = COD = AOD = 90.

To prove ABCD a square, we need to prove ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90.

Now, in AOB and COD

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

AOB = COD (Vertically opposite angles)

AOB COD (SAS congruence rule)

AB = CD (by CPCT) ... (1)

To prove ABCD a square, we need to prove ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90.

Now, in AOB and COD

AO = CO (Diagonals bisect each other)

OB = OD (Diagonals bisect each other)

AOB = COD (Vertically opposite angles)

AOB COD (SAS congruence rule)

AB = CD (by CPCT) ... (1)

And OAB = OCD (by CPCT)

But these are alternate interior angles for line AB and CD and alternate interior angle are equal to each other only when the two lines are parallel

AB || CD ... (2)

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

ABCD is a parallelogram

Now, in AOD and COD

AO = CO (Diagonals bisect each other)

AOD = COD (Given that each is 90)

OD = OD (common)

AOD COD (SAS congruence rule)

AD = DC ... (3)

But, AD = BC and AB = CD (opposite sides of parallelogram ABCD)

AB = BC = CD = DA

So, all the sides quadrilateral ABCD are equal to each other

Now, in ADC and BCD

AD = BC (Already proved)

AC = BD (given)

DC = CD (Common)

ADC BCD (SSS Congruence rule)

ADC = BCD (by CPCT)

But ADC + BCD = 180

ADC + ADC = 180

ADC = 180

ADC = 90

One of interior angle of ABCD quadrilateral is a right angle

Now, we have ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90. Therefore, ABCD is a square.

^{o}(co-interior angles)ADC + ADC = 180

^{o}ADC = 180

^{o}ADC = 90

^{o}One of interior angle of ABCD quadrilateral is a right angle

Now, we have ABCD is a parallelogram, AB = BC = CD = AD and one of its interior angle is 90. Therefore, ABCD is a square.

### Solution 6

(i) ABCD is a parallelogram.

DAC = BCA (Alternate interior angles) ... (1)

DAC = BCA (Alternate interior angles) ... (1)

And BAC = DCA (Alternate interior angles) ... (2)

But it is given that AC bisects A.

DAC = BAC ... (3)

But it is given that AC bisects A.

DAC = BAC ... (3)

From equations (1), (2) and (3), we have

DAC = BCA = BAC = DCA ... (4)

DCA = BCA

Hence, AC bisects C.

(ii)

DAC = BCA = BAC = DCA ... (4)

DCA = BCA

Hence, AC bisects C.

(ii)

From equation (4), we have

DAC = DCA

DA = DC (side opposite to equal angles are equal)

DAC = DCA

DA = DC (side opposite to equal angles are equal)

But DA = BC and AB = CD (opposite sides of parallelogram)

AB = BC = CD = DA

Hence, ABCD is rhombus

### Solution 7

Let us join AC

In ABC

BC = AB (side of a rhombus are equal to each other)

1 = 2 (angles opposite to equal sides of a triangle are equal)

In ABC

BC = AB (side of a rhombus are equal to each other)

1 = 2 (angles opposite to equal sides of a triangle are equal)

But 1 = 3 (alternate interior angles for parallel lines AB and CD)

2 = 3

So, AC bisects C.

Also, 2 = 4 (alternate interior angles for || lines BC and DA)

1 = 4

So, AC bisects A

Similarly, we can prove that BD bisects B and D as well.

### Solution 8

(i) Given that AC is bisector of A and C.

Or DAC = DCA

CD = DA (sides opposite to equal angles are also equal)

But DA = BC and AB = CD (opposite sides of rectangle are equal)

AB = BC = CD = DA

Or DAC = DCA

CD = DA (sides opposite to equal angles are also equal)

But DA = BC and AB = CD (opposite sides of rectangle are equal)

AB = BC = CD = DA

ABCD is a rectangle and all of its sides are equal.

Hence, ABCD is a square

Hence, ABCD is a square

(ii) Let us join BD

In BCD

BC = CD (side of a square are equal to each other)

CDB = CBD (angles opposite to equal sides are equal)

But CDB = ABD (alternate interior angles for AB || CD)

CBD = ABD

In BCD

BC = CD (side of a square are equal to each other)

CDB = CBD (angles opposite to equal sides are equal)

But CDB = ABD (alternate interior angles for AB || CD)

CBD = ABD

BD bisects B.

Also CBD = ADB (alternate interior angles for BC || AD)

CDB = ADB

BD bisects D.

### Solution 9

(i) In APD and CQB

ADP = CBQ (alternate interior angles for BC || AD)

AD = CB (opposite sides of parallelogram ABCD)

DP = BQ (given)

APD CQB (using SAS congruence rule)

(ii) As we had observed that APD CQB

AP = CQ (CPCT)

ADP = CBQ (alternate interior angles for BC || AD)

AD = CB (opposite sides of parallelogram ABCD)

DP = BQ (given)

APD CQB (using SAS congruence rule)

(ii) As we had observed that APD CQB

AP = CQ (CPCT)

(iii) In AQB and CPD

ABQ = CDP (alternate interior angles for AB || CD)

AB = CD (opposite sides of parallelogram ABCD)

BQ = DP (given)

AQB CPD (using SAS congruence rule)

(iv) As we had observed that AQB CPD

AQ = CP (CPCT)

ABQ = CDP (alternate interior angles for AB || CD)

AB = CD (opposite sides of parallelogram ABCD)

BQ = DP (given)

AQB CPD (using SAS congruence rule)

(iv) As we had observed that AQB CPD

AQ = CP (CPCT)

(v) From the result obtained in (ii) and (iv), we have

AQ = CP and AP = CQ

Since opposite sides in quadrilateral APCQ are equal to each other. So, APCQ is a

parallelogram.

### Solution 10

(i) In APB and CQD

APB = CQD (each 90

AB = CD (opposite sides of parallelogram ABCD)

ABP = CDQ (alternate interior angles for AB || CD)

APB CQD (by AAS congruency)

(ii) By using the result obtained as above

APB CQD, we have

AP = CQ (by CPCT)

APB = CQD (each 90

^{o})AB = CD (opposite sides of parallelogram ABCD)

ABP = CDQ (alternate interior angles for AB || CD)

APB CQD (by AAS congruency)

(ii) By using the result obtained as above

APB CQD, we have

AP = CQ (by CPCT)

### Solution 11

(i) Here AB = DE and AB || DE.

Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be

Now, if two opposite sides of a quadrilateral are equal and parallel to each other, it will be

a parallelogram.

Therefore, quadrilateral ABED is a parallelogram.

(ii) Again BC = EF and BC || EF.

Therefore, quadrilateral BEFC is a parallelogram.

Therefore, quadrilateral ABED is a parallelogram.

(ii) Again BC = EF and BC || EF.

Therefore, quadrilateral BEFC is a parallelogram.

(iii) Here ABED and BEFC are parallelograms.

AD = BE, and AD || BE

(Opposite sides of parallelogram are equal and parallel)

And BE = CF, and BE || CF

(Opposite sides of parallelogram are equal and parallel)

AD = CF, and AD || CF

(iv) Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and

AD = BE, and AD || BE

(Opposite sides of parallelogram are equal and parallel)

And BE = CF, and BE || CF

(Opposite sides of parallelogram are equal and parallel)

AD = CF, and AD || CF

(iv) Here one pair of opposite sides (AD and CF) of quadrilateral ACFD are equal and

parallel to each other,

so, it is a parallelogram.

(v) As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each

so, it is a parallelogram.

(v) As ACFD is a parallelogram, so, pair of opposite sides will be equal and parallel to each

other.

AC || DF and AC = DF

(vi) ABC and DEF.

AB = DE (given)

BC = EF (given)

AC = DF (ACFD is a parallelogram)

ABCDEF (by SSS congruence rule)

AC || DF and AC = DF

(vi) ABC and DEF.

AB = DE (given)

BC = EF (given)

AC = DF (ACFD is a parallelogram)

ABCDEF (by SSS congruence rule)

### Solution 12

Extend AB. Draw a line through C, which is parallel to AD, intersecting AE at point E.

Now, AECD is a parallelogram.

Now, AECD is a parallelogram.

(i) AD = CE (opposite sides of parallelogram AECD)

But AD = BC (given)

So, BC = CE

CEB = CBE (angle opposite to equal sides are also equal)

Now consider parallel lines AD and CE. AE is transversal line for them

A + CEB = 180 (angles on the same side of transversal)

A+ CBE = 180 (using the relationCEB = CBE) ... (1)

But B + CBE = 180 (linear pair angles) ... (2)

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

A = B

(ii) AB || CD

A + D = 180 (angles on the same side of transversal)

Also C + B = 180 (angles on the same side of transversal)

A + D = C + B

But AD = BC (given)

So, BC = CE

CEB = CBE (angle opposite to equal sides are also equal)

Now consider parallel lines AD and CE. AE is transversal line for them

A + CEB = 180 (angles on the same side of transversal)

A+ CBE = 180 (using the relationCEB = CBE) ... (1)

But B + CBE = 180 (linear pair angles) ... (2)

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

A = B

(ii) AB || CD

A + D = 180 (angles on the same side of transversal)

Also C + B = 180 (angles on the same side of transversal)

A + D = C + B

But A = B [using the result obtained proved in (i)]

C = D

(iii) In ABC and BAD

AB = BA (common side)

BC = AD (given)

B = A (proved before)

ABC BAD (SAS congruence rule)

(iv) ABCBAD

AC = BD (by CPCT)

## Quadrilaterals Exercise Ex. 8.2

### Solution 1

(i) In ADC, S and R are the mid points of sides AD and CD respectively.

In a triangle the line segment joining the mid points of any two sides of the triangle is

In a triangle the line segment joining the mid points of any two sides of the triangle is

parallel to the third side and is half of it.

SR || AC and SR = AC ... (1)

(ii) In ABC, P and Q are mid points of sides AB and BC respectively. So, by using

SR || AC and SR = AC ... (1)

(ii) In ABC, P and Q are mid points of sides AB and BC respectively. So, by using

mid-point theorem, we have

PQ || AC and PQ = AC ... (2)

Now using equations (1) and (2), we have

PQ || SR and PQ = SR ... (3)

PQ = SR

(iii) From equations (3), we have

PQ || SR and PQ = SR

Clearly one pair of opposite sides of quadrilateral PQRS is parallel and equal

Hence, PQRS is a parallelogram.

PQ || AC and PQ = AC ... (2)

Now using equations (1) and (2), we have

PQ || SR and PQ = SR ... (3)

PQ = SR

(iii) From equations (3), we have

PQ || SR and PQ = SR

Clearly one pair of opposite sides of quadrilateral PQRS is parallel and equal

Hence, PQRS is a parallelogram.

### Solution 2

In ABC, P and Q are mid points of sides AB and BC respectively.

PQ || AC and PQ = AC (using mid-point theorem) ... (1)

PQ || AC and PQ = AC (using mid-point theorem) ... (1)

In ADC

R and S are the mid points of CD and AD respectively

RS || AC and RS = AC (using mid-point theorem) ... (2)

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

PQ || RS and PQ = RS

As in quadrilateral PQRS one pair of opposite sides are equal and parallel to each other, so, it is a parallelogram.

Let diagonals of rhombus ABCD intersect each other at point O.

Now in quadrilateral OMQN

MQ || ON ( PQ || AC)

QN || OM ( QR || BD)

So, OMQN is parallelogram

MQN = NOM

PQR = NOM

But, NOM = 90

^{o}(diagonals of a rhombus are perpendicular to each other)

PQR = 90

^{o}

Clearly PQRS is a parallelogram having one of its interior angle as 90.

Hence, PQRS is rectangle.

### Solution 3

Let us join AC and BD

In ABC

P and Q are the mid-points of AB and BC respectively

PQ || AC and PQ = AC (mid point theorem) ... (1)

In ABC

P and Q are the mid-points of AB and BC respectively

PQ || AC and PQ = AC (mid point theorem) ... (1)

Similarly in ADC

SR || AC and SR = AC (mid point theorem) ... ... (2)

Clearly, PQ || SR and PQ = SR

As in quadrilateral PQRS one pair of opposite sides is equal and parallel to

each other, so, it is a parallelogram.

PS || QR and PS = QR (opposite sides of parallelogram)... (3)

Now, in BCD, Q and R are mid points of side BC and CD respectively.

QR || BD and QR = BD (mid point theorem) ... (4)

But diagonals of a rectangle are equal

AC = BD ... ... (5)

Now, by using equation (1), (2), (3), (4), (5) we can say that

PQ = QR = SR = PS

So, PQRS is a rhombus.

### Solution 4

By converse of mid-point theorem a line drawn, through the mid point of any side of a triangle and parallel to another side bisects the third side.

Now in ABD

EF || AB and E is mid-point of AD

So, this line will intersect BD at point G and G will be the mid-point of DB.

Now as EF || AB and AB || CD

EF || CD (Two lines parallel to a same line are parallel to each other)

Now in ABD

EF || AB and E is mid-point of AD

So, this line will intersect BD at point G and G will be the mid-point of DB.

Now as EF || AB and AB || CD

EF || CD (Two lines parallel to a same line are parallel to each other)

Now, in BCD, GF || CD and G is the midpoint of line BD. So, by using converse of mid-point theorem, F is the mid-point of BC.

### Solution 5

ABCD is a parallelogram

AB || CD

AB || CD

So, AE || FC

Again AB = CD (opposite sides of parallelogram ABCD)

AB = CD

AE = FC (E and F are midpoints of side AB and CD)

As in quadrilateral AECF one pair of opposite sides (AE and CF) are parallel and equal to each other. So, AECF is a parallelogram.

AF || EC (Opposite sides of a parallelogram)

Now, in DQC, F is mid point of side DC and FP || CQ (as AF || EC). So, by using converse of mid-point theorem, we can say that

P is the mid-point of DQ

DP = PQ ... (1)

Similarly, in APB, E is mid point of side AB and EQ || AP (as AF || EC). So, by using converse of mid-point theorem, we can say that

Q is the mid-point of PB

PQ = QB ... (2)

From equations (1) and (2), we may say that

DP = PQ = BQ

Hence, the line segments AF and EC trisect the diagonal BD.

### Solution 6

Let ABCD is a quadrilateral in which P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively.

Join PQ, QR, RS, SP and BD.

In ABD, S and P are mid points of AD and AB respectively.

So, By using mid-point theorem, we can say that

SP || BD and SP = BD ... (1)

Similarly in BCD

QR || BD and QR = BD ... (2)

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

SP || QR and SP = QR

As in quadrilateral SPQR one pair of opposite sides are equal and parallel to

each other.

So, SPQR is a parallelogram.

Join PQ, QR, RS, SP and BD.

In ABD, S and P are mid points of AD and AB respectively.

So, By using mid-point theorem, we can say that

SP || BD and SP = BD ... (1)

Similarly in BCD

QR || BD and QR = BD ... (2)

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

SP || QR and SP = QR

As in quadrilateral SPQR one pair of opposite sides are equal and parallel to

each other.

So, SPQR is a parallelogram.

Since, diagonals of a parallelogram bisect each other.

Hence, PR and QS bisect each other.

Hence, PR and QS bisect each other.

### Solution 7

(i) In ABC

Given that M is mid point of AB and MD || BC.

So, D is the mid-point of AC. (Converse of mid-point theorem)

(ii) As DM || CB and AC is a transversal line for them.

So, MDC + DCB = 180 (Co-interior angles)

MDC + 90 = 180

MDC = 90

MD AC

Given that M is mid point of AB and MD || BC.

So, D is the mid-point of AC. (Converse of mid-point theorem)

(ii) As DM || CB and AC is a transversal line for them.

So, MDC + DCB = 180 (Co-interior angles)

MDC + 90 = 180

MDC = 90

MD AC

(iii) Join MC

In AMD and CMD

AD = CD (D is the midpoint of side AC)

ADM = CDM (Each 90)

DM = DM (common)

AMDCMD (by SAS congruence rule)

AD = CD (D is the midpoint of side AC)

ADM = CDM (Each 90)

DM = DM (common)

AMDCMD (by SAS congruence rule)

So, AM = CM (by CPCT)

But AM = AB (M is mid point of AB)

So, CM = MA = AB

But AM = AB (M is mid point of AB)

So, CM = MA = AB