NCERT Solutions for Class 9

Mathematics
Chapter 8 – Quadrilaterals

1. The angles of quadrilateral are in the ratio 3:5:9:13 .Find all the angles of the
quadrilateral.

Ans:

Given: The quadrilateral angles are in the ratio 3:5:9:13.

To find: Angles of a quadrilateral

Consider's say the angle's common ratio is x .

Therefore, the angles will be 3x, 5x, 9x, and 13x respectively.

The sum of a quadrilateral's inner angles is 360 ,

 3x  5 x  9 x  13x  360
30 x  360
x  12

As a result, the angles can be calculated as

3 x  3  12  36
5 x  5  12  60
9 x  9  12  108
13 x  13  12  156

2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Ans:

Class IX Mathematics www.vedantu.com 1

Given: Diagonals of the parallelogram are the same.

To prove: It is a rectangle.

Consider ABCD be the given parallelogram.

Now we need to show that ABCD is a rectangle, by proving that one of its interior
angles is .

In ABC and DCB ,

AB = DC (side opposite to the parallelogram are equal)

BC = BC (in common)

AC = DB (Given)

ABC  DCB

(By SSS Congruence rule)

 ABC  DCB

The sum of the measurements of angles on the same side of a transversal is known
to be 180 o

.Hence, ABCD is a rectangle because it is a parallelogram with a 90 o inner angle.

3. Show that if the diagonals of a quadrilateral bisect each other at right angles,
then it is a rhombus.

Ans:

Given: The diagonals of a quadrilateral bisect at right angles.

To prove: It is a rhombus.

Consider ABCD be a quadrilateral, whose diagonals AC and BD bisect each other

at right angle

Class IX Mathematics www.vedantu.com 2

i.e.

OA  OC , OB  OD , and .

In order to prove ABCD a rhombus, we need to prove ABCD is the parallelogram

and all sides of the ABCD are the same.

In AOD

and COD ,

OA = OC (Diagonals of the parallelogram bisect each other)

AOD  COD

(Given)

OD = OD (Common)

AOD  COD

(By SAS congruence rule)

 AD  CD … (1)

Similarly, it can be proved that

AD  AB and CD  BC … (2)

From Equations (1) and (2),

AB  BC  CD  AD

To conclude, ABCD is a parallelogram because the opposite sides of the

quadrilateral ABCD are equal. ABCD is a rhombus because all of the sides of a
parallelogram ABCD are equal.

4. Show that the diagonals of a square are equal and bisect each other at right
angles.

Class IX Mathematics www.vedantu.com 3

Ans:

Given: A square is given.

To find: The diagonals of a square are the same and bisect each other at 90 o

Consider ABCD be a square.

Consider the diagonals AC and BD intersect each other at a point O.

We must first show that the diagonals of a square are equal and bisect each other at
right angles,

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

In ABC and DCB ,

AB  DC

(Sides of the square are equal)

ABC  DCB

(All the interior angles are of the value 90 o )

BC  CB

(Common side)

ABC  DCB

(By SAS congruency)

 AC  DB

(By CPCT)

Hence, the diagonals of a square are equal in length.

In AOB and COD ,

AOB  COD

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(Vertically opposite angles)

ABO  CDO

(Alternate interior angles)

AB = CD (Sides of a square are always equal)

AOB  COD

(By AAS congruence rule)

 AO  CO and OB  OD

(By CPCT)

As a result, the diagonals of a square are bisected.

In AOB and COB ,

Because we already established that diagonals intersect each other,

AO  CO

AB  CB

(Sides of a square are equal)

BO  BO

(Common)

AOB  COB

(By SSS congruency)

AOB  COB

(By CPCT)

However, (Linear pair)

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As a result, the diagonals of a square are at right angles to each other.

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

Ans:

Given: The diagonals of a quadrilateral are equal and bisect at 90 o

To find: It is a square

Consider the quadrilateral ABCD, in which the diagonals AC and BD cross at point
O.

The diagonals of ABCD are equal and bisect each other at right angles, which is a
given.

Therefore,

AC  BD, OA  OC , OB  OD , and .

We need to prove that ABCD is a parallelogram,

AB  BC  CD  AD

, and one of its interior angles being .

In AOB and COD , AO  CO

(Diagonals bisect each other)

OB  OD

(Diagonals bisect each other)

AOB  COD

(Vertically opposite angles)

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AOB  COD

(SAS congruence rule)

AB  CD (By CPCT) ... (1)

And,

OAB  OCD . (By CPCT)

However, for lines AB and CD, they are alternate interior angles, and alternate
interior angles are equal only when the two lines are parallel.

AB || CD ... (2)

From (1) and (2), we obtain

ABCD is a parallelogram.

In AOD , COD ,

AO  CO

(Diagonals bisect each other)

AOD  COD

(Given that each is )

OD  OD

(Common)

AOD  COD

(SAS congruence rule)

 AD  DC ... (3)

However,

AD  BC and AB  CD

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(Opposite sides of the parallelogram ABCD)

AB  BC  CD  DA

Therefore, all the sides of quadrilateral ABCD are equal to each other.

In ADC and BCD ,

AD  BC

(Already proved)

AC  BD

(Given)

DC  CD

(Common)

ADC  BCD

(SSS Congruence rule)

ADC  BCD

(By CPCT)

However,

ADC  BCD  180

(Co-interior angles)

ADC  ADC  180

2ADC  180
ADC  90

A right angle is one of the quadrilateral ABCD's inner angles.

As a result, we can see that ABCD is a parallelogram,

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AB  BC  CD  AD and one of its interior angles is .

Hence , ABCD is a square.

6. Diagonal AC of a parallelogram ABCD is bisecting A (see the given figure).

Show that

(i) It is bisecting C also,

(ii) ABCD is a rhombus

Ans:

Given: Diagonal AC of a parallelogram ABCD is bisecting A

To find: (i) It is bisecting C also,

(ii) ABCD is a rhombus

(i) ABCD is a parallelogram.

DAC  BCA

(Alternate interior angles) ... (1)

And,

BAC  DCA

(Alternate interior angles) ... (2)

However, it is given that AC is bisecting A

DAC  BAC ... (3)

From Equations (1), (2), and (3), we obtain

DAC  BCA  BAC  DCA ... (4)

DCA  BCA

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Hence, AC is bisecting C

(ii) From Equation (4), we obtain

DAC  DCA

DA  DC

(Side opposite to equal angles are equal)

However,

DA  BC and AB  CD

(Opposite sides of a parallelogram)

AB  BC  CD  DA

As a result, ABCD is a rhombus.

7. ABCD is a rhombus. Show that diagonal AC is bisecting A as well as C

and diagonal BD is bisecting B as well as D

Given: ABCD is a rhombus.

To find: Diagonal AC is bisecting A as well as C and diagonal BD is bisecting

B as well as D .

Ans:

Let us now join AC.

In ABC ,

BC  AB

(Sides of the rhombus are equal to each other)

1  2

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(Angles opposite to equal sides of a triangle are equal)

However,

1  3

(Parallel lines AB and CD have different interior angles.)

2  3

Hence, AC is bisecting C .

Also,

2  4

(Alternate interior angles for || lines BC and DA)

1  4

Hence, AC bisects A .

Also, it can be proved that BD is bisecting B and D as well.

8. ABCD is a rectangle in which diagonal AC bisects A as well as C . Show

that:

(i) ABCD is a square

Ans: It is given that ABCD is a rectangle.

A  C

1 1
 A  C (AC bisects A and C )
2 2

1
 DAC  DCA
2

CD = DA (Sides that are opposite to the equal angles are also equal)

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Also,

DA  BC and AB  CD

(Opposite sides of the rectangle are same)

AB  BC  CD  DA

ABCD is a rectangle with equal sides on all sides.

Hence, ABCD is a square.

(ii) Diagonal BD bisects B as well as D .

Ans:

Let us now join BD.

In BCD ,

BC  CD

(Sides of a square are equal to each other)

CDB  CBD

(Angles opposite to equal sides are equal)

However,

CDB  ABD

(Alternate interior angles for AB || CD )

CBD  ABD BD bisects B.

Also,

CBD  ADB

(Alternate interior angles for BC || AD )

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CDB  ABD BD bisects D and B

9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such

that DP = BQ (see the given figure). Show that:

(i) APD  CQB

Ans: (i) In APD and CQB ,

ADP  CBQ (Alternate interior angles for BC || AD )

AD  CB (Opposite sides of the parallelogram ABCD)

DP  BQ (Given)

APD  CQB (Using SAS congruence rule)

(ii) AP  CQ

Ans: As we had observed that APD  CQB ,

 AP  CQ (CPCT)

(iii) AQB  CPD

Ans: 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) AQ  CP

Ans: Since we had observed that

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AQB  CPD ,

 AQ  CP (CPCT)

(v) APCQ is a parallelogram

Ans: From the result obtained in (ii) and (iv),

AQ  CP and AP  CQ

APCQ is a parallelogram because the opposite sides of the quadrilateral are equal.

10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices

A and C on diagonal BD (See the given figure). Show that

(i) APB  CQD

Ans: In APB and CQD ,

APB  CQD (Each 90°)

AB  CD

(The opposite sides of a parallelogram ABCD)

ABP  CDQ (Alternate interior angles for AB || CD )

APB  CQD (By AAS congruency)

(ii) AP  CQ

Ans: By using

APB  CQD , we obtain

AP  CQ (By CPCT)

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11. In ABC and DEF , AB  DE , AB || DE , BC  EF and BC || EF . Vertices A, B
and C are joined to vertices D, E and F respectively (see the given figure). Show
that

(i) Quadrilateral ABED is a parallelogram

Ans: It is given that AB  DE and AB || DE .

A parallelogram is formed when two opposite sides of a quadrilateral are equal and
parallel to one other.

As a result, ABED is a parallelogram in quadrilateral form.

(ii) Quadrilateral BEFC is a parallelogram

Ans: Again, BC = EF and BC || EF

Therefore, quadrilateral BCEF is a parallelogram.

(iii) AD || CF and AD = CF

Ans: As we had observed that ABED and BEFC are parallelograms, therefore

AD = BE and AD || BE (Opposite sides of a parallelogram are equal and parallel)

And, BE = CF and BE || CF (Opposite sides of a parallelogram are equal and

parallel)

 AD  CF and AD || CF

(iv) Quadrilateral ACFD is a parallelogram

Ans: We know it's a parallelogram since one set of opposing sides (AD and CF) of
quadrilateral ACFD are equal and parallel to one other.

(v) AC = DF

Ans: Due to the fact that ACFD is a parallelogram, the pair of opposing sides will
be equal and parallel to one another.

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 AC || DF and AC  DF

(vi) ABC  DEF

Ans: In ABC and DEF ,

AB  DE (Given)

BC  EF (Given)

AC  DF (ACFD is a parallelogram)

ABC  DEF (By SSS congruence rule)

12. ABCD is a trapezium in which AB || CD and AD = BC (see the given figure).

Show that

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB

produced at E.]

(i) ∠A = ∠B

Ans: AD  CE (Opposite sides of parallelogram AECD)

However,

AD  BC (Given)

Therefore,

BC  CE

CEB  CBE (Angle opposite to the equal sides are also equal)

Consideing parallel lines AD and CE. AE is the transversal line for them.

(Angles on a same side of transversal)

(Using the relation ∠CEB = ∠CBE) ... (1)

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However, (Linear pair angles) ... (2)

From Equations (1) and (2), we obtain

A  B

(ii) ∠C = ∠D

Ans: AB || CD (Angles on a same side of the transversal)

Also,

C  B  180 (Angles on a same side of a transversal)

A  D  C  B

However,

A  B [Using the result obtained in (i)]

C  D

(iii) ABC  BAD

Ans: In ABC and BAD ,

AB  BA (Common side)

BC  AD (Given)

B  A (Proved before)

ABC  BAD (SAS congruence rule)

(iv) diagonal AC = diagonal BD

Ans: We had seen that,

ABC  BAD

 AC  BD (By CPCT)

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Exercise - 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides

AB, BC, CD and DA (see the given figure). AC is a diagonal. Show that:

1
(i) SR || AC and SR  AC
2

Ans: In ADC , S and R are the mid-points of sides AD and CD respectively.

In a triangle, the line segment connecting the midpoints of any two sides is parallel
to and half of the third side.

1
 SR || AC and SR  AC ... (1)
2

(ii) PQ = SR

Ans: In ∆ABC, P and Q are mid-points of sides AB and BC respectively. Therefore,

by using midpoint theorem,

1
PQ || AC and PQ  AC ... (2)
2

Using Equations (1) and (2), we obtain

1
PQ || SR and PQ  SR ... (3)
2

 PQ  SR

(iii) PQRS is a parallelogram.

Ans: From Equation (3), we obtained

PQ || SR and PQ  SR

Clearly, one pair of quadrilateral PQRS opposing sides is parallel and equal.

PQRS is thus a parallelogram.

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2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB,
BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Ans:

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

To find: Quadrilateral PQRS is a rectangle

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

1
PQ || AC , PQ  AC (Using mid-point theorem) ... (1)
2

In ADC ,

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

1
RS || AC , RS  AC (Using mid-point theorem) ... (2)
2

From Equations (1) and (2), we obtain

PQ || RS and PQ  RS

It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is

equal and parallel to each other. At position O, the diagonals of rhombus ABCD
should cross.

In quadrilateral OMQN,

MQ ON ( PQ AC )

QN OM ( QR BD)

Hence , OMQN is a parallelogram.

MQN  NOM
PQR  NOM

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Since,

NOM  90 (Diagonals of the rhombus are perpendicular to each other)

PQR  90

Clearly, PQRS is a parallelogram having one of its interior angles as .

So , PQRS is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC,
CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Ans:

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.

Let us join AC and BD.

In ABC ,

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

1
 PQ || AC and PQ  AC (Mid-point theorem) ... (1)
2

Similarly in ADC ,

1
SR || AC , SR  AC (Mid-point theorem) ... (2)
2

Clearly,

PQ || SR and PQ  SR

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It is a parallelogram because one pair of opposing sides of quadrilateral PQRS is
equal and parallel to each other.

 PS || QR , PS  QR (Opposite sides of parallelogram) ... (3)

In BCD , Q and R are the mid-points of side BC and CD respectively.

1
 QR || BD , QR  BD (Mid-point theorem) ... (4)
2

Also, the diagonals of a rectangle are equal.

 AC  BD …(5)

By using Equations (1), (2), (3), (4), and (5), we obtain

PQ  QR  SR  PS

So , PQRS is a rhombus

4. ABCD is a trapezium in which AB || 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
the given figure). Show that F is the mid-point of BC.

Ans:

Given: ABCD is a trapezium in which AB || 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.

To prove: F is the mid-point of BC.

Let EF intersect DB at G.

We know that a line traced through the mid-point of any side of a triangle and
parallel to another side bisects the third side by the reverse of the mid-point theorem.

In ABD ,

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

Hence , G will be the mid-point of DB.

As

EF AB , AB CD ,

 EF || CD (Two lines parallel to the same line are parallel)

In BCD ,

GF || CD and G is the mid-point of line BD. So , by using converse of mid-point

theorem, F is the mid-point of BC.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD

respectively (see the given figure). Show that the line segments AF and EC
trisect the diagonal BD.

Ans:

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.

ABCD is a parallelogram.

AB || CD

And hence,

AE || FC

Again, AB = CD (Opposite sides of parallelogram ABCD)

1 1
AB  CD
2 2

AE  FC (E and F are mid-points of side AB and CD)

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In quadrilateral AECF, one pair of the opposite sides (AE and CF) is parallel and
same to each other. So , AECF is a parallelogram.

 AF || EC (Opposite sides of a parallelogram)

In DQC , F is the mid-point of side DC and

FP || CQ (as AF || EC ).

So , by using the converse of mid-point theorem, it can be said that P is the mid-
point of DQ.

 DP  PQ ... (1)

Similarly, in

APB , E is the mid-point of side AB and

EQ || AP (as AF || EC ).

As a result, the reverse of the mid-point theorem may be used to say that Q is the
mid-point of PB.

 PQ  QB ... (2)

From Equations (1) and (2),

DP  PQ  BQ

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

6. Show that the line segments joining the mid-points of the opposite sides of a
quadrilateral bisect each other.

Ans:

Given: ABCD is a quadrilateral

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

Let ABCD be a quadrilateral in which the midpoints of sides AB, BC, CD, and DA
are P, Q, R, and S, respectively. PQ, QR, RS, SP, and BD are all joined

S and P are the mid-points of AD and AB, respectively, in ABD

. As a result, using the mid-point theorem, it can be shown that

1
SP || BD , SP  BD ... (1)
2

Similarly in

BCD ,

1
QR || BD , QR  BD ... (2)
2

From the Equations (1) and (2), we obtain

SP || QR , SP  QR

One set of opposing sides of quadrilateral SPQR is equal and parallel to each other.
SPQR is a parallelogram as a result.

The diagonals of a parallelogram are known to bisect each other.

As a result, PR and QS cut each other in half.

7. 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. Show that

(i) D is the mid-point of AC

Ans: In ABC ,

It is given that M is the mid-point of AB and

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MD || BC .

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

(ii) MD  AC

Ans: As DM || CB and AC is a transversal line for them, therefore,

(Co-interior angles)

1
(iii) CM  MA  AB
2

Ans: Join MC.

In AMD and CMD ,

AD  CD (D is the mid-point of side AC)

ADM  CDM (Each )

DM = DM (Common)

AMD  CMD (By SAS congruence rule)

Therefore,

AM  CM (By CPCT)

However,

1
AM  AB (M is mid-point of AB)
2

Therefore, it is said that

1
CM  AM  AB
2

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