# Class 9 NCERT Solutions Maths Chapter 8 - Quadrilaterals

Ex. 8.1

Ex. 8.2

## Quadrilaterals Exercise Ex. 8.1

### Solution 1

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 2

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 3

(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 4

(i)

Given:

DAC = CAB ... (1)

and DCA = BCA ... (2)

Now, AD || BC and AC is a transversal

DAC = BCA

CAB = BCA

In ΔABC, we have

CAB = BCA

BC = AB ... (Sides opposite to equal angles are equal)

But BC = AD and AB = CD ... Since ABCD is a rectangle.

AB = BC = AD = CD

All the sides of this rectangle are equal.

Thus, ABCD is a square.

(ii)

Since, ABCD is a square and we know that the diagonals of a square bisect its angles.

BD bisects B as well asD.

### Solution 5

(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)

AQBCPD (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)

AQBCPD (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 6

(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 7

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

(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