# NCERT Solutions for Class 9 Maths Chapter 8 Quadrilaterals

NCERT solutions for class 9 maths chapter 8 Quadrilaterals: A figure designed through four straight lines is called Quadrilaterals. It has four sides and four angles and the sum of all the four angles is always 360 degrees. There are many types of quadrilaterals which you will study in this particular chapter. Solutions of NCERT class 9 maths chapter 8 Quadrilaterals is covering the problems related to all the concepts in a very comprehensive manner.

The topic is easy to relate to real-life because many shapes around us are in the form of quadrilaterals such as top of a table, paper, wall, roof, etc. To deal with problems related to this chapter, the awareness about the properties of different types of quadrilaterals is necessary. CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals covers all the questions to give you guidance while solving the questions. In this particular chapter, you will get to know about angles sum property, rhombus, parallelogram, square, rectangle, trapezium, kite, etc. The chapter is starting with recollecting the angle sum property of a quadrilateral. In total there are 2 exercises that consist of 37 questions. NCERT solutions for class 9 maths chapter 8 Quadrilaterals has the solutions to all the 37 questions. If I talk apart from this chapter, then NCERT solutions for other classes and subjects can also be downloaded using the link given.

## NCERT solutions for class 9 maths chapter 8 Quadrilaterals Excercise: 8.1

Given :  The angles of a quadrilateral are in the ratio  $\small 3:5:9:13$
Let the angles of quadrilateral be   $3x,5x,9x ,13x$.

Sum of all angles is 360.

Thus,   $3x+5x+9x +13x=360 \degree$

$\Rightarrow 30x=360 \degree$

$\Rightarrow x=12 \degree$

All four  angles are : $3x=3\times 12=36\degree$

$5x=5\times 12=60 \degree$

$9x=9\times 12=108 \degree$

$13x=13\times 12=156 \degree$

Given: ABCD is a parallelogram with AC=BD.

To prove: ABCD is a rectangle.

Proof : In $\triangle$ABC and $\triangle$BAD,

BC= AD            (Opposite sides of parallelogram)

AC=BD              (Given )

AB=AB               (common)

$\triangle$ABC $\cong$ $\triangle$BAD   (By SSS)

$\angle ABC=\angle BAD$             (CPCT)

and     $\angle ABC+\angle BAD=180 \degree$      (co - interior angles)

$2\angle BAD=180 \degree$

$\angle BAD= 90 \degree$

Hence, it is a rectangle.

Given: the diagonals of a quadrilateral bisect each other at right angles i.e.BO=OD.

To prove: ABCD is a rhombus.

Proof : In $\triangle$AOB and $\triangle$AOD,

$\angle AOB=\angle AOD$        (Each $90 \degree$)

BO=OD              (Given )

AO=AO               (common)

$\triangle$AOB $\cong$ $\triangle$AOD   (By SAS)

Similarly, AB=BC and BC=CD

Hence, it is a rhombus.

Given : ABCD is a square i.e. AB=BC=CD=DA.

To prove : the diagonals of a square are equal and bisect each other at right angles i.e. AC=BD,AO=CO,BO=DO and $\angle COD=90 \degree$

Proof : In $\triangle$BAD and $\triangle$ABC,

$\angle BAD = \angle ABC$        (Each $90 \degree$)

AB=AB               (common)

$\triangle$BAD $\cong$ $\triangle$ABC   (By SAS)

BD=AC           (CPCT)

In $\triangle$AOB and $\triangle$COD,

$\angle$OAB=$\angle$OCD        (Alternate angles)

AB=CD               (Given )

$\angle$OBA=$\angle$ODC       (Alternate angles)

$\triangle$AOB $\cong$ $\triangle$COD   (By AAS)

AO=OC ,BO=OD           (CPCT)

In $\triangle$AOB and $\triangle$AOD,

OB=OD        (proved above)

OA=OA          (COMMON)

$\triangle$AOB $\cong$ $\triangle$AOD   (By SSS)

$\angle$AOB=$\angle$AOD           (CPCT)

$\angle$AOB+$\angle$AOD =$180 \degree$

2. $\angle$AOB = $180 \degree$

$\angle$AOB =$90 \degree$

Hence,  the diagonals of a square are equal and bisect each other at right angles.

Given : ABCD is a quadrilateral with  AC=BD,AO=CO,BO=DO,$\angle$COD =$90 \degree$

To prove: ABCD is a square.

Proof: Since the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a rhombus.

Thus, AB=BC=CD=DA

In $\triangle$BAD and $\triangle$ABC,

AB=AB               (common)

BD=AC

$\triangle$BAD $\cong$ $\triangle$ABC   (By SSS)

$\angle BAD = \angle ABC$    (CPCT)

$\angle$BAD+$\angle$ABC =$180 \degree$           (Co-interior angles)

2. $\angle$ABC = $180 \degree$

$\angle$ABC =$90 \degree$

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

it bisects $\small \angle C$ also.

Given: $\angle$DAC=$\angle$BAC ................1

$\angle$DAC=$\angle$BCA.................2  (Alternate angles)

$\angle$BAC=$\angle$ACD .................3  (Alternate angles)

From equation  1,2 and 3, we get

$\angle$ACD=$\angle$BCA...................4

Hence, diagonal AC bisect angle C also.

ABCD is a rhombus.

Given: $\angle$DAC=$\angle$BAC ................1

$\angle$DAC=$\angle$BCA.................2  (Alternate angles)

$\angle$BAC=$\angle$ACD .................3  (Alternate angles)

From equation  1,2 and 3, we get

$\angle$ACD=$\angle$BCA...................4

From 2 and 4, we get

$\angle$ACD=$\angle$DAC

In $\triangle$ ADC,

$\angle$ACD=$\angle$DAC       (proved above )

AD=DC         (In a triangle,sides opposite to equal angle are equal)

A parallelogram whose adjacent sides are equal , is a rhombus.

Thus, ABCD is a rhombus.

In $\triangle$ ADC,

AD = CD          (ABCD is a rhombus)

$\angle$3=$\angle$1.................1(angles opposite to equal sides are equal )

$\angle$3=$\angle$2.................2 (alternate angles)

From 1 and 2, we have

$\angle$1=$\angle$2.................3

and        $\angle$1=$\angle$4.................4 (alternate angles)

From 1 and 4, we get

$\angle$3=$\angle$4.................5

Hence, from 3 and 5, diagonal AC bisects angles A as well as angle C.

In $\triangle$ ADB,

AD = AB          (ABCD is a rhombus)

$\angle$5=$\angle$7.................6(angles opposite to equal sides are equal )

$\angle$7=$\angle$6.................7 (alternate angles)

From 6 and 7, we have

$\angle$5=$\angle$6.................8

and        $\angle$5=$\angle$8.................9(alternate angles)

From 6 and 9, we get

$\angle$7=$\angle$8.................10

Hence, from 8 and 10, diagonal BD bisects angles B as well as angle D.

Given: ABCD is a rectangle with AB=CD and BC=AD $\angle$1=$\angle$2 and $\angle$3=$\angle$4.

To prove: ABCD is a square.

Proof :     $\angle$1=$\angle$4 .............1(alternate angles)

$\angle$3=$\angle$4 ................2(given )

From 1 and 2, $\angle$1=$\angle$3.....................................3

In $\triangle$ADC,

$\angle$1=$\angle$3          (from  3 )

DC=AD             (In a triangle, sides opposite to equal angle are equal )

A rectangle whose adjacent sides are equal is a square.

Hence, ABCD is a square.

diagonal BD bisects $\small \angle B$ as well as $\small \angle D$.

In $\triangle$ ADB,

AD = AB          (ABCD is a square)

$\angle$5=$\angle$7.................1(angles opposite to equal sides are equal )

$\angle$5=$\angle$8.................2 (alternate angles)

From 1 and 2, we have

$\angle$7=$\angle$8.................3

and        $\angle$7 =$\angle$6.................4(alternate angles)

From 1 and 4, we get

$\angle$5=$\angle$6.................5

Hence, from 3 and 5, diagonal BD bisects angles B as well as angle D.

Given:   In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$.

To prove :$\small \Delta APD\cong \Delta CQB$

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ       (Given )

$\angle$ADP=$\angle$CBQ     (alternate angles)

AD=BC       (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$                (By SAS)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$.

To prove :$\small AP=CQ$

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ       (Given )

$\angle$ADP=$\angle$CBQ     (alternate angles)

AD=BC       (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$                (By SAS)

$\small AP=CQ$      (CPCT)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$.

To prove :$\small \Delta AQB\cong \Delta CPD$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ       (Given )

$\angle$ABQ=$\angle$CDP     (alternate angles)

AB=CD       (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$                (By SAS)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$.

To prove :$\small AQ=CP$

Proof :

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ       (Given )

$\angle$ABQ=$\angle$CDP     (alternate angles)

AB=CD       (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$                (By SAS)

$\small AQ=CP$                       (CPCT)

Given: In parallelogram ABCD, two points P and Q are taken on diagonal BD such that $\small DP=BQ$.

To prove: APCQ is a parallelogram

Proof :

In $\small \Delta APD\, \, and\, \, \Delta CQB,$

DP=BQ       (Given )

$\angle$ADP=$\angle$CBQ     (alternate angles)

AD=BC       (opposite sides of a parallelogram)

$\small \Delta APD\cong \Delta CQB$                (By SAS)

$\small AP=CQ$       (CPCT)...............................................................1

Also,

In $\small \Delta AQB\, \, and\, \, \Delta CPD,$

DP=BQ       (Given )

$\angle$ABQ=$\angle$CDP     (alternate angles)

AB=CD       (opposite sides of a parallelogram)

$\small \Delta AQB\cong \Delta CPD$                (By SAS)

$\small AQ=CP$                       (CPCT)........................................2

From equation 1 and 2, we get

$\small AP=CQ$

$\small AQ=CP$

Thus, opposite sides of quadrilateral APCQ are equal so APCQ is a parallelogram.

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A  and C on diagonal BD.

To prove : $\small \Delta APB\cong \Delta CQD$

Proof: In$\small \Delta APB\, \, and\, \, \Delta CQD$,

$\angle$APB=$\angle$CQD             (Each $90 \degree$)

$\angle$ABP=$\angle$CDQ            (Alternate angles)

AB=CD                (Opposite sides of a parallelogram )

Thus, $\small \Delta APB\cong \Delta CQD$               (By SAS)

Given: ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A  and C on diagonal BD.

To prove : $\small AP=CQ$

Proof: In$\small \Delta APB\, \, and\, \, \Delta CQD$,

$\angle$APB=$\angle$CQD             (Each $90 \degree$)

$\angle$ABP=$\angle$CDQ            (Alternate angles)

AB=CD                (Opposite sides of a parallelogram )

Thus, $\small \Delta APB\cong \Delta CQD$               (By SAS)

$\small AP=CQ$                 (CPCT)

Given :  In $\small \Delta ABC$ and $\small \Delta DEF$,  $\small AB=DE,AB\parallel DE,BC=EF$ and    $\small BC\parallel EF$.

To prove :  quadrilateral ABED is a parallelogram

Proof : In ABED,

AB=DE          (Given)

AB||DE            (Given )

Hence, quadrilateral ABED is a parallelogram.

Given: In $\small \Delta ABC$ and $\small \Delta DEF$,  $\small AB=DE,AB\parallel DE,BC=EF$ and                              $\small BC\parallel EF$.

To prove: quadrilateral BEFC is a parallelogram

Proof: In BEFC,

BC=EF          (Given)

BC||EF            (Given )

Hence, quadrilateral BEFC is a parallelogram.

To prove: $\small AD\parallel CF$ and $\small AD=CF$

Proof :

In ABED,

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

From 1 and 3, we get

To prove :  quadrilateral ACFD is a parallelogram

Proof :

In ABED,

In BEFC,

BE=CF.................3(BEFC is a parallelogram)

BE||CF .................4(BEFC is a parallelogram)

From 2 and 4 , we get

From 1 and 3, we get

From 5 and 6, we get

Thus, quadrilateral ACFD is a parallelogram

In ACFD,

AC=DF     (Since, ACFD is a parallelogram which is prooved in part (iv) of the question)

In $\small \Delta ABC$ and $\small \Delta DEF$,

AB=DE                   (Given )

BC=EF                    (Given )

AC=DF                ( proved in (v) part)

$\small \Delta ABC\cong \Delta DEF$        (By SSS rule)

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Given: ABCD is a trapezium in which  $\small AB\parallel CD$ and  $\small AD=BC$

To prove :$\small \angle A=\angle B$

Proof:      Let $\angle$ A be $\angle$1, $\angle$ABC be $\angle$2, $\angle$ EBC be $\angle$3, $\angle$ BEC be $\angle$4.

In AECD,

AE||DC             (Given)

Hence, AECD is a parallelogram.

From 1 and 2, we get

CE=BC

In $\triangle$BCE,

$\angle 3=\angle 4$.................3  (opposite angles of equal sides)

$\angle 2+\angle 3=180 \degree$...................4(linear pairs)

$\angle 1+\angle 4=180 \degree$ .....................5(Co-interior angles)

From 4 and 5, we get

$\angle 2+\angle 3=\angle 1+\angle 4$

$\therefore \angle 2=\angle 1 \rightarrow \angle B=\angle A$          (Since,$\angle 3=\angle 4$)

Given: ABCD is a trapezium in which  $\small AB\parallel CD$ and  $\small AD=BC$

To prove :$\small \angle C=\angle D$

Proof:   Let $\angle$ A be $\angle$1, $\angle$ABC be $\angle$2, $\angle$ EBC be $\angle$3, $\angle$ BEC be $\angle$4.

$\angle 1+\angle D= 180 \degree$       (Co-interior angles)

$\angle 2+\angle C= 180 \degree$            (Co-interior angles)

$\therefore \angle 1+\angle D=\angle 2+\angle C$

Thus, $\small \angle C=\angle D$                (Since ,$\small \angle 1=\angle 2$ )

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Given: ABCD is a trapezium in which  $\small AB\parallel CD$ and  $\small AD=BC$

To prove :$\small \Delta ABC\cong \Delta BAD$

Proof: In  $\small \Delta ABC\, \, and\, \, \, \Delta BAD$,

AB=AB            (Common )

$\angle ABC=\angle BAD$              (proved in (i)  )

Thus, $\small \Delta ABC\cong \Delta BAD$              (By SAS rule)

[Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

Given: ABCD is a trapezium in which  $\small AB\parallel CD$ and  $\small AD=BC$

To prove: diagonal AC $\small =$ diagonal BD

Proof: In  $\small \Delta ABC\, \, and\, \, \, \Delta BAD$,

AB=AB            (Common )

$\angle ABC=\angle BAD$              (proved in (i)  )

Thus, $\small \Delta ABC\cong \Delta BAD$              (By SAS rule)

diagonal AC $\small =$ diagonal BD            (CPCT)

## NCERT solutions for class 9 maths chapter 8 Quadrilaterals Excercise: 8.2

Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$). AC is a diagonal.

To prove :  $\small SR\parallel AC$  and   $\small SR=\frac{1}{2}AC$

Proof: In $\triangle$ACD,

S is the midpoint of DA.                (Given)

R  is the midpoint of DC.               (Given)

By midpoint theorem,

$\small SR\parallel AC$  and   $\small SR=\frac{1}{2}AC$

Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$). AC is a diagonal.

To prove : $\small PQ=SR$

Proof : In $\triangle$ACD,

S is mid point of DA.                (Given)

R  is mid point of DC.               (Given)

By mid point theorem,

$\small SR\parallel AC$  and   $\small SR=\frac{1}{2}AC$...................................1

In $\triangle$ABC,

P is mid point of AB.                (Given)

Q  is mid point of BC.               (Given)

By mid point theorem,

$\small PQ\parallel AC$  and   $\small PQ=\frac{1}{2}AC$.................................2

From 1 and 2,we get

$\small PQ\parallel SR$          and   $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$

Given :   ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig $\small 8.29$). AC is a diagonal.

To prove : PQRS is a parallelogram.

Proof : In PQRS,

Since,

$\small PQ\parallel SR$      and       $\small PQ=SR$.

So,PQRS is a parallelogram.

Given: ABCD is a rhombus in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA . AC, BD are  diagonals.

To prove: the quadrilateral PQRS is a rectangle.

Proof: In $\triangle$ACD,

S is midpoint of DA.                (Given)

R  is midpoint of DC.               (Given)

By midpoint theorem,

$\small SR\parallel AC$  and   $\small SR=\frac{1}{2}AC$...................................1

In $\triangle$ABC,

P is midpoint of AB.                (Given)

Q  is mid point of BC.               (Given)

By mid point theorem,

$\small PQ\parallel AC$  and   $\small PQ=\frac{1}{2}AC$.................................2

From 1 and 2,we get

$\small PQ\parallel SR$          and   $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$     and $\small PQ\parallel SR$

So,the quadrilateral PQRS is a parallelogram.

Similarly, in $\triangle$BCD,

Q is mid point of BC.                (Given)

R  is mid point of DC.               (Given)

By mid point theorem,

$\small QR\parallel BD$

So,  QN || LM ...........5

LQ || MN ..........6  (Since, PQ || AC)

From 5 and 6, we get

LMPQ is a parallelogram.

Hence, $\small \angle$LMN=$\small \angle$LQN   (opposite angles of the parallelogram)

But, $\small \angle$LMN= 90     (Diagonals of a rhombus are perpendicular)

so,   $\small \angle$LQN=90

Thus, a parallelogram whose one angle is right angle,ia a rectangle.Hence,PQRS is a rectangle.

Given: ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively.

To prove: the quadrilateral PQRS is a rhombus.

Proof :

In $\triangle$ACD,

S is the midpoint of DA.                (Given)

R  is the midpoint of DC.               (Given)

By midpoint theorem,

$\small SR\parallel AC$  and   $\small SR=\frac{1}{2}AC$...................................1

In $\triangle$ABC,

P is the midpoint of AB.                (Given)

Q  is the midpoint of BC.               (Given)

By midpoint theorem,

$\small PQ\parallel AC$  and   $\small PQ=\frac{1}{2}AC$.................................2

From 1 and 2, we get

$\small PQ\parallel SR$          and   $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$     and $\small PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram.

Similarly, in $\triangle$BCD,

Q is the midpoint of BC.                (Given)

R  is the midpoint of DC.               (Given)

By midpoint theorem,

$\small QR\parallel BD$                  and      $\small QR=\frac{1}{2}BD$...................5

AC = BD.......................6(diagonals )

From 2,  5 and 6, we get

PQ=QR

Thus, a parallelogram whose adjacent sides are equal is a rhombus. Hence, PQRS is a rhombus.

Given: ABCD is a trapezium in which  $\small AB\parallel DC$, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. $\small 8.30$).

To prove: F is the mid-point of BC.

In $\triangle$ABD,

E is the midpoint of AD.                (Given)

EG || AB                          (Given)

By converse of  midpoint theorem,

G is the midpoint of BD.

In $\triangle$BCD,

G is mid point of BD.                (Proved above)

FG || DC                          (Given)

By converse of  midpoint theorem,

F is the midpoint of BC.

Given: In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively

To prove: the line segments AF and EC trisect the diagonal BD.

AB=CD                             (Given)

$\frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AE=CF$         (E and F are midpoints of AB and CD)

AE=CF                             (Given)

AE || CF               (Opposite sides of a parallelogram)

Hence,  AECF is a parallelogram.

In $\triangle$ DCQ,

F is the midpoint of DC.      (given )

FP || CQ           (AECF  is a parallelogram)

By converse of midpoint theorem,

P is the mid point of DQ.

DP= PQ....................1

Similarly,

In $\triangle$ ABP,

E is the midpoint of AB.      (given )

EQ || AP          (AECF  is a parallelogram)

By converse of midpoint theorem,

Q is the midpoint of PB.

OQ= QB....................2

From 1 and 2, we have

DP = PQ = QB.

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

Given: ABCD is a  quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA. AC, BD are diagonals.

To prove: the line segments joining the mid-points of the opposite sides of a quadrilateral bisect each other.

Proof: In $\triangle$ACD,

S is the midpoint of DA.                (Given)

R  is midpoint of DC.               (Given)

By midpoint theorem,

$\small SR\parallel AC$  and   $\small SR=\frac{1}{2}AC$...................................1

In $\triangle$ABC,

P is the midpoint of AB.                (Given)

Q  is the midpoint of BC.               (Given)

By midpoint theorem,

$\small PQ\parallel AC$  and   $\small PQ=\frac{1}{2}AC$.................................2

From 1 and 2, we get

$\small PQ\parallel SR$          and   $\small PQ=SR=\frac{1}{2}AC$

Thus, $\small PQ=SR$     and $\small PQ\parallel SR$

So, the quadrilateral PQRS is a parallelogram and diagonals of a parallelogram bisect each other.

Thus, SQ and PR bisect each other.

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse    AB and parallel to BC intersects AC at D.

To prove :D is mid point of AC.

Proof: In $\triangle$ABC,

M is mid point of AB.                (Given)

DM || BC            (Given)

By converse of mid point theorem,

D is the mid point of AC.

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersects AC at D.

To prove : $\small MD\perp AC$

Proof :  $\angle$ADM = $\angle$ACB      (Corresponding angles)

$\angle$ADM= $90 \degree$.           ($\angle$ACB = $90 \degree$)

Hence,$\small MD\perp AC$.

Given: ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.

To prove :  $\small CM=MA=\frac{1}{2}AB$

Proof : In $\triangle$ABC,

M is the midpoint of AB.                (Given)

DM || BC            (Given)

By converse of midpoint theorem,

D is the midpoint of AC  i.e.  AD = DC.

In $\triangle$ AMD  and $\triangle$ CMD,

$\angle$ADM = $\angle$ CDM    (Each  right angle)

DM = DM      (Common)

$\triangle$ AMD $\cong$$\triangle$ CMD      (By SAS)

AM = CM         (CPCT)

But ,$\small AM=\frac{1}{2}AB$

Hence,$\small CM=MA=\frac{1}{2}AB$.

## NCERT solutions for class 9 maths chapter wise

 Chapter No. Chapter Name Chapter 1 NCERT solutions for class 9 maths chapter 1 Number Systems Chapter 2 CBSE NCERT solutions for class 9 maths chapter 2 Polynomials Chapter 3 Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry Chapter 4 NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables Chapter 5 CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry Chapter 6 Solutions of NCERT class 9 maths chapter 6 Lines And Angles Chapter 7 NCERT solutions for class 9 maths chapter 7 Triangles Chapter 8 CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals Chapter 9 Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles Chapter 10 NCERT solutions for class 9 maths chapter 10 Circles Chapter 11 CBSE NCERT solutions for class 9 maths chapter 11 Constructions Chapter 12 Solutions of NCERT class 9 maths chapter 12 Heron’s Formula Chapter 13 NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes Chapter 14 CBSE NCERT solutions for class 9 maths chapter 14 Statistics Chapter 15 Solutions of NCERT class 9 maths chapter 15 Probability

## How to use NCERT solutions for class 9 maths chapter 8 Quadrilaterals

• Read and memorize the properties related to all kinds of quadrilaterals.

• Have a glance through some solved to understand the pattern of the solution to that particular question.

• Now check your learning on the practice exercises problems.

• If you stuck in any question then you can assist yourself using NCERT solutions for class 9 maths chapter 8 Quadrilaterals.

• Following the above-written steps, you can get 100% out of the chapter.