Lesson: Quadrilaterals

Exercise 8.1 (12)

Question: 1

The angles of quadrilateral are in the ratio

3: 5: 9:13

.

Find all the angles of the quadrilateral.

Solution:

Let ‘

a’

be the common ratio between the angles.

Then the angles of the quadrilateral are

and 3 , 5 , 9 13 .a a a a

Sum of the interior angles of a quadrilateral

360

Then,

3 5 9 13 360

30 360

12

a a a a

a

a

Hence, the angles of the quadrilateral are:

3 3 12 36

5 5 12 60

9 9 12 108

a

a

a

and

13 13 12 156a

Question: 2

If the diagonals of a parallelogram are equal, then

show that it is a rectangle.

Solution:

Let

ABCD

be the parallelogram and

AC BD

.

We need to prove that

ABCD

is a rectangle.

In

ΔABC

and

ΔBAD

,

AB BA

(Common)

BC AD

(Opposite sides of a parallelogram are

equal)

AC BD

(Given)

Therefore,

ABC BAD

(By

SSS

congruence condition)

A B

(By CPCT)

Also,

180A B

(Sum of the angles on the

same side of the transversal)

2 180

90

A

A

Since

ABCD

is a parallelogram with one angle equal

to

90

ABCD

is a rectangle.

Hence, it is proved.

Question: 3

Show that if the diagonals of a quadrilateral bisect

each other at right angles, then it is a rhombus.

Solution:

Let

ABCD

be a quadrilateral whose diagonals

AC

and

BD

bisect each other at

O

.

Then,

,OA OC OB OD

and

90 .AOB BOC COD DOA

We need to show that ABCD is a rhombus.

AB BC CD AD

Now, in

AOB

and

,COB

OA OC

(Given)

AOB COB

90

(Given)

OB OB

(Common)

Therefore,

AOB COB

by

SAS

congruence condition.

Thus,

AB BC

(By CPCT)

Similarly, we can prove that

AB BC CD AD

Thus, ABCD is a rhombus.

Question: 4

Show that the diagonals of a square are equal and

bisect each other at right angles.

Solution:

Let

ABCD

be a square and its diagonals

AC

and

BD

intersect each other at

O

.

We need to show that the diagonals of a square

ABCD

are equal and bisect each other at right angles.

i.e.,

, ,AC BD AO OC OB OD

and

90AOB

Now, in

ABC

and

,BAD

BA AB

(Common)

90ABC BAD

BC AD

(

ABCD

is a square)

∴

ABC BAD

by

SAS

congruence rule.

Thus,

AC BD

by

CPCT

Hence, the diagonals are equal.

Now, in

AOB

and

COD

,

BAO DCO

(Alternate interior angles)

AOB COD

(Vertically opposite angles)

AB CD

(Sides of the square)

Therefore,

AOB COD

by AAS congruence condition.

Thus,

AO CO

and

OB OD

by CPCT. (Diagonal bisects

each other.)

Now, in

AOB

and

COB

,

OB OB

(Common)

AO CO

(As proved above)

AB CB

(Sides of the square)

Therefore,

AOB COB

by

SSS

congruence

condition.

AOB COB

But

180AOB COB

(Linear pair)

Thus,

90AOB COB

Similarly, we can show that,

90DOA COD

Hence, it is proved that diagonals bisect each other

at right angles.

Question: 5

Show that if the diagonals of a quadrilateral are

equal and bisect each other at right angles, then it is

a square.

Solution:

Let

ABCD

be a quadrilateral in which diagonal

AC

and

BD

are equal and bisect each other at right

angle

s

.

We need to prove that quadrilateral

ABCD

is a

square.

Now, In

AOB

and

COD

,

AO CO

(

Diagonals bisect each other)

AOB COD

(Right angles)

OB OD

(

Diagonals bisect each other)

Therefore,

AOB COD

by

SAS

congruence

condition.

Thus,

AB CD

(By

CPCT ...

….. (i)

Also,

OAB OCD

(By

CPCT

)

||AB CD

(

Alternate interiors angles are equal)

Similarly, we can prove that

.AOD COB

Thus,

AD BC

by CPCT …………. (ii)

And

||AD BC

Thus,

ABCD

is a parallelogram.

Now, in

AOD

and

COD

,

AO CO

(

Diagonals bisect each other)

AOD COD

(Right angles)

OD OD

(Common)

Therefore,

AOD COD

by SAS congruence rule.

Thus,

AD CD

(By CPCT) --- (iii)

Now using (i), (ii) and (iii), we get

AD BC CD AB

Hence,

ABCD

is parallelogram with all sides equal.

Now, let us consider

ΔACD

and

ΔBDC

.

AC BD

(Given)

DC DC

(Common)

AD BC

(As proved above)

So, by

SSS

congruence rule,

Δ ΔACD BDC

ADC BCD

by CPCT.

But,

180ADC BCD

(Co-interior angles)

2 180

90

ADC

ADC

Thus, ABCD is a parallelogram with one angle equal

to

90

.

ABCD

is a rectangle. But,

ABCD

has all equal

sides.

Hence, the given quadrilateral

ABCD

is a square.

Question: 6

Diagonal

AC

of a parallelogram

ABCD

bisects

A

(See Fig. 8.19).

Show that

(i) it bisects

C

also,

(ii)

ABCD

is a rhombus.

Solution:

(i)To show:

AC

bisects

C

.

AB DC

(Opposite sides of

gm

)

DAC ACB

and

CAB ACD

But,

DAC CAB

ACB ACD

AC

bisects

C

(ii) In

,ACD

ACD CAB

(Alternate interior angle)

But,

CAB CAD

(

AC

bisects angle

A

)

ACD CAD

AD CD

(Opposite sides of equal angles)

Also,

AB CD

and

BC DA

(Opposite sides of a

|| gm

)

So,

AB CD AD BC

Thus,

ABCD

is a rhombus.

Question: 7

ABCD is a rhombus. Show that diagonal

AC

bisects

A

as well as

C

and diagonal

BD

bisects

B

as

well as

D

.

Solution:

Since

ABCD

is a rhombus,

So,

AB BC CD DA

In ,ABC

AB BC

BCA BAC

… (i)

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

Also,

||AB CD

(

ABCD

is a rhombus)

DAC BCA

(Alternate interior angles)

BAC DAC

(From (i))

Therefore,

AC

bisects

.A

Similarly, the diagonal

AC

bisects

C

.

Also, by above method we can also prove that

diagonal

BD

bisects

B

as well as

.D

Hence, it is proved.

Question: 8

ABCD

is a rectangle in which diagonal

AC

bisects

A

as well as

C

. Show that:

(i)

ABCD

is a square

(ii) Diagonal

BD

bisects

B

as well as

D

.

Solution:

(i)

Since

ABCD

is a rectangle,

||AD BC

DCA BAC

(Alternate interior angle)

But,

DCA ACB

(

AC

bisects

C

)

BAC ACB

Now, in

,ABC

BAC ACB AB BC

(Sides opposite to equal

angles of a triangle are equal)

Also,

CD AB

(Opposite sides of a rectangle)

Therefore,

AB BC CD AD

.

Thus,

ABCD

is a square.

(ii)

In Δ

BCD

,

BC CD

(As proved above)

CDB CBD

(Angles opposite to equal sides

are equal).

Also,

CDB ABD

(Alternate interior angles)

CBD ABD

Thus,

BD

bisects

B

.

Now,

CBD ADB

(Alternate interior angle)

( )CDB ADB CDB CBD

Thus,

BD

bisects

D

.

Question: 9

In parallelogram

ABCD

, two points

P

and

Q

are

taken on diagonal

BD

such that

DP BQ

(See Fig.

8.20). Show that:

(i)

APD CQB

(ii)

AP CQ

(iii)

AQB CPD

(iv)

AQ CP

(v)

APCQ

is a parallelogram

ABQ CDP

(Alternate interior angles)

AB CD

(Opposite sides of a

|| gm

)

Thus,

AQB CPD

by

SAS

congruence rule.

(iv)

AQ CP

by

CPCT

(Using

AQB CPD

from part (iii))

(v) From (ii) and (iv),

APCQ

has opposite sides equal.

A quadrilateral is called a parallelogram if its

opposite sides are equal. Thus,

APCQ

is a

parallelogram.

Question: 10

ABCD

is a parallelogram and

AP

and

CQ

are

perpendiculars from vertices

A

and

C

on diagonal

BD

(see Fig. 8.21).

Show that

(i)

APB CQD

(ii)

AP CQ

Question: 11

In and ||, , ,ABC 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 Fig. 8.22).

Show that

(i) Quadrilateral

ABED

is a parallelogram

(ii) Quadrilateral

BEFC

is a parallelogram

(iii)

||AD CF

and

AD CF

(iv) quadrilateral

ACFD

is a parallelogram

(v)

AC DF

(vi)

. ABC DEF

Solution:

(i) In quadrilateral

ABED,

AB DE

and

||AB DE

(Given)

So, the quadrilateral

ABED

is a parallelogram.

(ii) In quadrilateral

BEFC,

BC EF

and

||BC EF

(Given)

So, the quadrilateral

BEFC

is a parallelogram.

(iii) Since

ABED

is a parallelogram,

AD BE

and

...AD BE i

(Opposite sides of a parallelogram)

Also,

BEFC

is a parallelogram.

BE CF

and

...BE CF ii

(Opposite sides of a parallelogram)

From Eq. (i) and (ii),

AD CF

and

| .|AD CF

(iv) From part (iii), we have

AD CF

and

| .|AD CF

Thus,

ACFD

is a parallelogram.

(v) Since

ACFD

is a parallelogram (From part (iv)),

AD BC

(See Fig. 8.23).

Show that

(i)

A B

(ii)

C D

(iii)

ABC BAD

(iv)

Diagonal iagonal AC D BD

[Hint: Extend

AB

and draw a line through

C

parallel

to

DA

intersecting

AB

at

E.

]

Solution:

Let us draw a line through

C

parallel to

DA

intersecting

AB

at

E

when

AB

is extended.

(i) In

,ADCE

AD CE

(By construction)

and

AE DC

(∵

AB CD

is given).

Thus,

ADCE

is a parallelogram.

CE AD

(Opposite sides of a parallelogram)

But,

AD BC

(Given)

BC CE

So, in

∆BCE

,

CBE CEB

(Angle opposite to equal sides)

also,

180A CEB

.

(

ADCE

is a parallelogram)

⇒

180A CBE

…. (1) (

CBE CEB

)

but,

180ABC CBE

…(2) (Linear pair)

Comparing (1) & (2), we get

A ABC

(ii) In

,ABCD AB CD

(Given)

180A D B C

(Angles on the same sides of transversal)

( )A D A C A B

D C

(iii) InΔ

ABC

and

ΔBAD

,

AB BA

(Common)

DAB CBA

(As proved above in (i))

AD BC

(Given)

Thus,

ABC BAD

(By

SAS

congruence condition)

(iv)

AC BD

by

CPCT

as

.ABC BAD

Also,

AC

&

BD

are diagonals of

ABCD

.

Exercise 8.2 (7)

Question: 1

ABCD

is a quadrilateral in which

P

,

Q

,

R

and

S

are

midpoints of the sides

AB, BC, CD

and

DA

(See Fig

8.29).

AC

is a diagonal. Show that:

(i)

||SR AC

and

1

2

SR AC

(ii)

PQ SR

(iii)

PQRS

is a parallelogram.

Solution:

(i) In

ΔDAC

,

R

and

S

are the midpoints of

DC

and

DA

respectively.

Thus, by the midpoint theorem,

||SR AC

and

1

2

SR AC

(ii) In

ΔBAC,

P

and

Q

are the midpoints of

AB

and

BC

respectively.

Thus by the midpoint theorem,

||PQ AC

and

1

2

PQ AC

also,

1

2

SR AC

(Using part(i)).

Thus,

PQ SR

(iii)

||SR AC

from (i)

and,

||PQ AC

from (ii)

||SR PQ

from (i) and (ii)

also

,

PQ SR

(from part (ii))

Thus,

PQRS

is a parallelogram.

Question: 2

ABCD

is a rhombus and

P, Q, R

and

S

are the

midpoints of the sides

AB, BC, CD

and

DA

respectively. Show that the quadrilateral

PQRS

is a

rectangle.

Solution:

Let us join

AC

and

BD

.

In

ΔDRS

and

ΔBPQ,

DS BQ

1

(

1

2 2

)AD BC AD BC

SDR QBP

( )ADC CBA

1 1

2 2

( )DR BP DC AB DC AB

Thus,

DRS BPQ

by

SAS

congruence condition.

RS PQ

by

CPCT

--- (i)

Similarly,

QCR SAP

by

SAS

congruence condition.

RQ SP

by

CPCT

--- (ii)

Now, in

ΔABC,

P

and

Q

are the midpoints of

AB

and

BC

respectively.

So, by the midpoint theorem,

... | |PQ AC iii

Also, in

∆ADC,

R

and

S

are the midpoints of

DC

and

AD

respectively.

So, by the midpoint theorem,

||SR AC

||PQ SR

(Using (iii))

Thus,

PQRS

is a parallelogram.

Consider

,OMQN

( )OM QN BD QR

and

( )MQ ON PQ AC

OMQN

is a parallelogram.

MON MQN

(Opposite angles of a

parallelogram)

But,

90MON

(

Diagonals of a rhombus are perpendicular to each

other)

90

90

MQN

PQR

Hence,

PQRS

is a parallelogram with one right angle.

Thus,

PQRS

is a rectangle.

Question: 3

ABCD

is a rectangle and

P, Q, R

and

S

are midpoints

of the sides

AB, BC, CD

and

DA

respectively. Show

that the quadrilateral

PQRS

is a rhombus.

Solution:

Let us join

AC

and

BD.

We need to show that

PQRS

is a rhombus.

Now, in

ADB

,

P

and

S

are the midpoints of

AB

and

AD

respectively.

Thus,

||PS BD

and

...

1

2

PS BD i

(Using the midpoint theorem)

In

,BDC

R

and

Q

are the midpoints of

DC

and

BC

respectively.

||RQ BD

and

1

2

RQ BD

(Midpoint theorem) …(ii)

From Eq. (i) and (ii),

||PS RQ

and

PS RQ

Thus,

PQRS

is a parallelogram.

||PQ SR

and

PQ SR

(Opposite sides of parallelogram) --- (iii)

Now, in

ABC

,

P

and

Q

are the midpoints of side

AB

and

BC

respectively.

Thus,

||PQ AC

and

1

2

PQ AC

(Midpoint theorem) --

- (iv)

Now,

AC BD

(Diagonals of a rectangle are equal) -

-- (v)

From equations (i), (ii), (iii), (iv) and (v), we get

PQ QR SR PS

So,

PQRS

is a rhombus.

Question: 4

ABCD

is a trapezium in which

,||AB DC BD

is a

diagonal and

E

is the midpoint of

AD

. A line is

drawn through

E

parallel to

AB

intersecting

BC

at

F

(See Fig. 8.30). Show that

F

is the midpoint of

BC

.

Solution:

Let

EF

intersects

BD

at

G

.

Now, in

ΔBAD

,

E

is the midpoint of

AD

(Given)

and

|| ( .)EG AB EF AB

So,

G

is the midpoint of

BD

(Converse of the

midpoint theorem).

Now, in

ΔBDC

,

G

is the midpoint of

BD

(As proved above)

and

|| ( )GF DC EF AB DC

Hence,

F

is the midpoint of

BC

(Converse of the

midpoint theorem).

Question: 5

In a parallelogram

ABCD

,

E

and

F

are the midpoints

of sides

AB

and

CD

respectively (See Fig. 8.31). Show

that the line segments

AF

and

EC

trisect the

diagonal

BD

.

Solution:

E

and

F

are the midpoints of sides

AB

and

CD

respectively in a parallelogram

ABCD

.

We need to show that

.DP PQ QB

Since in the quadrilateral

AECF

,

|| ( )AE FC AB CD

and

)(

1 1

2 2

AE FC AB CD AB CD AE FC

Thus,

AECF

is a parallelogram.

||AF EC

and

AF EC

(Opposite sides of a

parallelogram).

Now, in

ΔDQC

,

F

is midpoint of the side

DC (

Given)

and

.|| ( || )FP CQ AF EC

P

is the midpoint of

DQ

(Converse of the midpoint

theorem)

DP PQ i

Similarly, in ∆

APB

,

E

is midpoint of side

AB

(Given)

and

.|| || ) (EQ AP AF EC

Q

is the midpoint of

PB

(Converse of the midpoint

theorem)

PQ QB ii

From equations (i) and (ii),

DP PQ BQ

Hence, the line segments

AF

and

EC

trisect the

diagonal

BD

.

Question: 6

Show that the line segments joining the mid-points

of the opposite sides of a quadrilateral bisect each

other.

Solution:

Let

PQRS

be a quadrilateral and

A, B, C

and

D

are

the midpoints of

PQ, QR, RS

and

SP

respectively.

We need to show that

AC

and

BD

bisect each other.

Now, in

ΔPSQ,

D

and

A

are the midpoints of

PS

and

PQ

respectively.

Thus,

1||DA SQ

(Using the midpoint theorem)

Similarly, in

ΔSRQ,

C

and

B

are the midpoints of

SR

and

RQ

respectively.

Thus,

| 2|BC SQ

From (1) and (2),

||DA BC

Similarly, we can show that

||DC AB

.

Thus,

ABCD

is parallelogram.

AC

and

BD

are the diagonals of the parallelogram

ABCD

. So, they will bisect each other.

Question: 7

ABC

is a triangle right angled at

C

.

A

line through

the midpoint

M

of hypotenuse

AB

and parallel to

BC

intersects

AC

at

D

. Show that

(i)

D

is the midpoint of

AC

(ii)

MD AC

(iii)

1

2

CM MA AB

Solution:

(i) In

ΔACB

,

90C

.

Since

M

is the midpoint of

AB

and

||MD BC

,

by the converse of the midpoint theorem,

D

is the midpoint of

AC.

CD DA

(ii)

||MD BC

90ADM ACB

(Corresponding angles).

Hence,

.MD AC

(iii) In

ΔAMD

and

ΔCMD,

AD CD

(As proved above)

ADM CDM

(Each 90°)

DM = DM

(Common)

Thus,

AMD CMD

(By

SAS

congruence

condition)

AM CM

by

CPCT

Also,

1

2

AM AB

(

M

is midpoint of

AB

)

Hence,

1

2

CM MA AB

.