Lesson: Triangles

Exercise 7.1 (8)

Question: 1

In quadrilateral ACBD,

AC AD

and AB bisects

A

(See Fig. 7.16). Show that

ABC ABD  

. What

can you say about BC and BD?

Solution:

Given:

In quadrilateralACBD,

AC AD

and AB bisects

A

We need to prove that

ABC ABD  

Now, in

ABC

and

ABD

AB AB

(Common)

CAB DAB  

(AB is the angle bisector)

AC AD

(Given)

Therefore,

ABC ABD  

by SAS congruence

condition.

BC BD

(By CPCT)

Question: 2

ABCD is a quadrilateral in which AD=BC and

DAB CBA  

(see Fig7.17). Prove that

(i)

ABD BAC  

(ii)

BD AC

(iii)

.ABD BAC  

Solution:

Given:

AD BC

and

DAB CBA  

(i)In

ABD

and

,BAC

AB BA

(Common)

DAB CBA  

(Given)

AD BC

(Given)

Therefore,

ABD BAC  

by SAS congruence

condition.

(ii) Since,

ABD BAC  

Therefore, BD=AC(By CPCT)

(iii) Since,

ABD BAC  

Therefore,

ABD BAC  

(By CPCT)

Question: 3

AD and BC are equal perpendiculars to a line segment

AB (see Fig. 7.18). Show that CD bisects AB.

Solution:

Given:

AD AB

and

.BC AB

To Prove:

OA OB

Proof:

In and ,AOD BOC 

A B  

(As

AD AB

and

BC AB

)

AOD BOC  

(Vertically opposite angles)

AD=BC (Given)

Therefore,

AOD BOC  

(By AAS congruence

condition)

This implies,

AO OB

(By CPCT)

or CD bisects AB.

Question: 4

l and m are two parallel lines intersected by another

pair of parallel lines p and q (See Fig. 7.19). Show that

.ABC CDA  

Solution:

Given:

l m

and

p q

To prove:

ABC CDA  

Proof:

ABC

and

,CDA

BCA DAC  

(Alternate interior angles)

AC CA

(Common)

BAC DCA  

(Alternate interior angles)

Therefore,

ABC CDA  

(By ASA congruence

condition.)

Question: 5

Line l is the bisector of angle

A

and B is any point

on l. BP and BQ are perpendiculars from B to the arms

A

(see Fig. 7.20). Show that:

(i)

APB AQB  

(ii)BP=BQ or B is equidistant from the arms of

.A

Solution:

Given:

l is the bisector of angle

A

BP and BQ are perpendiculars on arms of

A

(i)

In and ,APB AQB 

P Q  

(Right angles)

BAP BAQ  

(l is the bisector)

AB AB

(Common)

Therefore,

APB AQB  

by AAS congruence

condition.

(ii)

BP BQ

by CPCT. Therefore, B is equidistant

from the arms of

A

Question: 6

In Fig. 7.21,

,AC AE AB AD 

and

BAD EAC  

Show that

.BC DE

Solution:

Given:

,AC AE AB AD 

and

BAD EAC  

We need to prove that

BC DE

Now,

BAD EAC  

(Given)

Adding

DAC

on both the sides, we get

  BAD DAC EAC DAC    

BAC EAD   

ABC

and

,ADE

AC AE

(Given)

BAC EAD  

(As proved above)

AB AD

(Given)

Therefore,

ABC ADE  

by SAS congruence condition.

BC DE

(By CPCT).

Question: 7

AB is a line segment and P is its midpoint. D and E are

points on the same side of AB such that

BAD ABE  

and

EPA DPB  

(See Fig. 7.22).

Show that:

(i)

DAP EBP  

(ii)

AD BE

Solution:

Given:

P is the midpoint of AB. i.e.,

.AP PB

BAD ABE  

and

EPA DPB  

…(i)

Since,

EPA DPB  

Adding

DPE

both the sides

 EPA DPE DPB DPE     

DPA EPB   

Now, In

DAP

and

   ,EBP DPA EPB

(As proved above)

AP BP

(Given)

BAD ABE  

(Given)

Therefore,

DAP EBP  

by ASA Congruence condition.

(ii)

AD BE

by CPCT.

Question: 8

In right triangle ABC, right angled at C, M is the

midpoint of hypotenuse AB.C is joined to M and

produced to a point D such that

.DM CM

Point D

is joined to point B. Show that:

(i)

AMC BMD  

(ii)

DBC

is a right angle.

(iii)

DBC ACB  

(iv)

CM = AB

Solution:

Given,

90 ,MC  

is the midpoint of AB i.e.,

AM MB

and

DM CM

(i)In

AMC

and

,BMD

AM BM

(Given)

CMA DMB  

(Vertically opposite angles)

CM DM

(Given)

Therefore,

ΔAMC ΔBMD

by SAS congruence

condition.

 

i ..AC BD 

(By CPCT)

(ii)

ACM BDM  

(By CPCT)

Therefore,

||AC BD

as alternate interior angles are

equal.

Now,

  180ACB DBC   

(Co-interiors angles)

90  180DBC   

90 DBC  

(iii)In

DBC

and

,ACB

BC=CB (Common)

ACB DBC  

(Right angles)

DB AC

(Using (i))

Therefore,

DBC ACB  

by SAS congruence condition.

 

iiDC AB 

(By CPCT)

(iv)DC=AB (Using (ii))

DM CM AB  

 

AsCM CM AB DM CM   

CM AB

 

Exercise 7.2 (8)

Question: 1

In an isosceles triangle ABC, with

,AB AC

the

bisectors of

B

and

C

intersect each other at O.

Join A to O. Show that:

(i)

OB OC

(ii)AO bisects

A

Solution:

Given,

,AB AC

the bisectors of

B

and

C

intersect each

other at O.

(i) Since, ABC is an isosceles with

,AB AC

B C   

1 1

2 2

B C   

(As OB and OC are angle bisectors)

OBC OCB   

OB OC 

(Side opposite to the equal angles are

equal.)

(ii)

In and ,AOB AOC 

AB AC

(Given)

AO AO

(Common)

OB OC

(As above proved)

Therefore,

AOB AOC  

by SSS congruence

condition.

BAO CAO  

By CPCT

Thus, AO bisects

A

Question: 2

,ABC AD

is the perpendicular bisector of BC (see

Fig. 7.30). Show that

ABC

is an isosceles triangle in

which

.AB AC

Solution:

Given,

AD is the perpendicular bisector of BC.

To show,

AB AC

Proof,

ADB

and

,ADC

AD AD

(Common)

ADB ADC  

BD CD

(AD is the perpendicular bisector)

Therefore,

ADB ADC  

by SAS congruence

condition.

AB=AC (By CPCT).

Question: 3

ABC

is an isosceles triangle in which altitudes BE

and CF are drawn to equal sides AC and AB

respectively (see Fig. 7.31). Show that these altitudes

are equal.

Solution:

Given,

in ABC

BE and CF are altitudes.

AC AB

To show,

BE=CF

Proof:

In and ,AEB AFC 

A A  

(Common)

AEB AFC  

(Right angles)

AB AC

(Given)

Therefore,

AEB AFC  

by AAS congruence

condition.

Thus,

BE CF

by CPCT.

Question: 4

∆ABC is a triangle in which altitudes BE and CF to

sides AC and AB are equal (See Fig. 7.32). Show that:

(i)

ABE ACF  

(ii)

,AB AC

i.e., ABC is an isosceles triangle.

Solution:

Given,

  , &ABC BE CF

are two altitudes such that,

BE CF

(i)

In and ,ABE ACF 

A A  

(Common)

AEB AFC  

(Right angles)

BE=CF(Given)

Therefore,

ABE ACF  

by AAS congruence

condition.

(ii) Thus,

AB AC

by CPCT and therefore ABC is an

isosceles triangle.

Question: 5

andABC DBC 

are two isosceles triangles on the

same base BC (See Fig. 7.33). Show that

.ABD ACD  

Solution:

Given,

ABC

and

DBC

are two isosceles triangles.

To show:

.ABD ACD  

Proof:

Let us join AD

ABD

and

,ACD

AD AD

(Common)

AB AC

(

ABC

is an isosceles triangle.)

BD CD

(

BCD

is an isosceles triangle.)

Therefore,

ABD ACD  

(By SSS congruence condition).

Thus,

ABD ACD  

by CPCT.

Question: 6

ΔABC is an isosceles triangle in which

AB AC

Side BA is produced to D such that

AD AB

(See Fig.

7.34). Show that

BCD

is a right angle.

Solution:

Given,

AB AC

and

AD AB

AC AD 

To show:

BCD

is a right angle.

Proof:

In ,ABC

AB AC

(Given)

 

ACB ABC i    

(Angles opposite to the equal

sides are equal.)

In ,ACD

AC AD

(Given)

 

CDA DCA ii    

(Angles opposite to the equal

sides are equal.)

Now,

In ,BCD

180

180 

DBC BCD CDB

ABC BCD CDA

      

       

180ACB BCD DCA       

(From (i) and (ii))

  180

180

2 18

( )

ACB DCA BCD

BCD BCD

BCD

       

     

   

Hence, it is proved.

Question: 7

ABC is a right angled triangle in which

90A  

and

.AB AC

Find

B

and

C

Solution:

Given,

90A  

andAB AC

B C   

(Angles opposite to the equal sides are

equal.)

Now,

 180A B C     

(Sum of the interior angles of a

triangle is 180



90  2 180

2 90 

45 

    

   

Thus,

45B C    

Question: 8

Show that the angles of an equilateral triangle are

60

each.

Solution:

Let ABC be an equilateral triangle.

BC AC AB 

(Length of all sides is same)

A B C     

(Angles opposite to the equal sides

are equal.)

Also,

180

3 180

A B C

      

   

Therefore,

60A B C      

Thus, the angles of an equilateral triangle are

60

each.

Exercise 7.3 (5)

Question: 1

ABC

and DBC

are two isosceles triangles on the

same base BC and vertices A and D are on the same

side of BC (see Fig. 7.39). If AD is extended to BC

such that it intersects BC at P, show that:

(i)

ABD ACD  

(ii)

ABP ACP  

(iii)AP bisects

A

as well as

.D

(iv)AP is the perpendicular bisector of BC.

Solution:

Given,

andABC DBC 

are two isosceles triangles.

(i)

In and ,ABD ACD 

AD=AD (Common)

AB=AC (

ABC

is isosceles)

BD=CD (

DBC

is isosceles)

Therefore,

ABD ACD  

by SSS congruence

condition.

(ii)

In and ,ABP ACP 

AP=AP (Common)

PAB PAC  

ABD ACD  

So, by CPCT

AB AC

(

ABC

is isosceles)

Therefore,

ABP ACP  

by SAS congruence condition.

(iii)

by   .PAB PAC CPCT as ABD ACD     

Therefore, AP bisects

 

    1.A

Also,

In and ,BPD CPD 

PD=PD (Common)

BD=CD (

DBC

is isosceles.)

soby .( )BP CP ABP ACP CPCT   

Therefore,

BPD CPD  

by SSS congruence condition.

Thus,

 

     by  2.BDP CDP CPCT

By(1) and (2), we can say that AP bisects

A

as well

D

 

 y asB ( )iv BPD CPD CPCT BPD CPD

BP CP

     



Proved above

Also,

180BPD CPD   

(BC is a straight line.)

   

   

2 180

BPD

Therefore,

AP is the perpendicular bisector of BC.

Question: 2

AD is an altitude of an isosceles triangle ABC in

which

AB AC

. Show that:

(i)AD bisects BC

(ii)AD bisects

A

Solution:

Given,

AD is an altitude and

AB AC

(i) In

and ,ABD ACD 

90ADB ADC    

AB AC

(Given)

AD AD

(Common)

Therefore,

ABD ACD  

by RHS congruence condition.

Now,

BD CD

(By CPCT)

Thus, AD bisects BC

(ii)

BAD CAD  

(By CPCT)

Thus, AD bisects

A

Question: 3

Two sides AB and BC and median AM of one triangle

ABC are respectively equal to sides PQ and QR and

median PN of

PQR

(See Fig. 7.40). Show that:

(i)

ABM PQN  

(ii)

ABC PQR  

Solution:

Given,

,AB PQ BC QR 

and

AM PN

(i)

BC BM

and

QR QN

(AM and PN are medians)

also,

BC QR

(Given)

1 1



BC QR 

BM QN 

ABM

and

,PQN

AM PN

(Given)

AB PQ

(Given)

BM QN

(As proved above)

Therefore,

ABM PQN  

by SSS congruence

condition.

(ii)In

ABC

and

,PQR

AB PQ

(Given)

ABC PQR  

by CPCT,

Since,

ABM PQN  

BC QR

(Given)

Therefore,

ABC PQR  

by SAS congruence

condition.

Question: 4

BE and CF are two equal altitudes of a triangle ABC.

Using RHS congruence rule, prove that the triangle

ABC is isosceles.

Solution:

Given,

BE CF

and

BE AC

and

.CF AB

To show triangle ABC is an isosceles triangle.

Now,

BEC

and

,CFB

90BEC CFB    

(As

BE AC

and

CF AB

)

BC CB

(Common)

BE CF

(Given)

Therefore,

BEC CFB  

by RHS congruence

condition.

ECB FBC   

(By CPCT)

ACB ABC   

Thus,

,ABC AB AC 

(As sides opposite to the equal angles are equal.)

Thus, ΔABC is an isosceles triangle.

Question: 5

ABC is an isosceles triangle with

AB AC

Draw

.AP BC

Show that

.B C  

Solution:

Given, ABC is an isosceles triangle with

,AB AC

and

.AP BC

ABP

and

,ACP

90APB APC    

(Given)

AB AC

(Given)

AP AP

(Common)

Therefore,

ABP ACP  

by RHS congruence

condition.

Thus,

B C  

(By CPCT).

Exercise 7.4 (6)

Question: 1

Show that in a right angled triangle, the hypotenuse is

the longest side.

Solution:

Let us consider

ABC

be a triangle right angled at A.

Now,

    180A B C    

(The sum of all the angles of a

triangle is 180°)

as is90  90 .B C A     

 A

is the largest angle of the triangle, then the side

opposite to it must be the longest line segment.

So, BC is the longest side in

ABC

Also, BC is the hypotenuse. (Opposite side to the right

angle)

Therefore, BC is the hypotenuse which is the longest

side of the right-angled triangle ABC.

Question: 2

In Fig. 7.48, sides AB and AC of

ABC

are extended

to points P and Q respectively.

Also,

.PBC QCB  

Show that

.AC AB

Solution:

Given, sides AB and AC of

ABC

are extended to

points P and Q respectively.

Also,

PBC QCB  

Since, AP is a line segment,

 



180

180  ..

ABC PBC

ABC PBC i

     

      

Also,

Since, AQ is also a line segment,

 

180

180 

ACB QCB

ACB QCB ii

     

      

Now,

 



Given

180 180

PBC QCB

  

       

ABC ACB   

   

 

Byusing  &i ii

Thus,

 ,AC AB

as side opposite to the greater angle is

longer.

Question: 3

InFig. and 

Showthat

7.49, .

B A C D

AD BC

     



Solution:

Given,

and .B A C D     

In ,OAB

Since,

B A  

(Given)

Therefore,

 

AO BO i   

(Side opposite to the

smaller angle is smaller)

Again,

In ,OCD

Since,

C D  

(Given)

Therefore,

 

OD OC ii   

(Side opposite to the greater angle is longer)

Adding (i) and (ii),

AO OD BO OC

AD BC

  

 

Henceproved.

Question: 4

AB and CD are respectively the smallest and longest

sides of a quadrilateral ABCD (See Fig. 7.50).

Show that

and .A C B D     

Solution:

Given AB and CD are respectively the smallest and

longest sides of a quadrilateral ABCD.

,ABD

Since, AB is the smallest side of quadrilateral ABCD.

 

AB AD ADB ABD i        

(Angle opposite to the longer side is larger.)

Similarly,

In ,BCD

Since, CD is the largest side ofquadrilateral ABCD.

 BC DC

 

BDC CBD ii      

(Angle opposite to the longer side is larger.)

Adding (i) and (ii) we get,

 



ADB BDC ABD CBD

ADC ABC

D B

B D

    

   

   

Similarly,

In ,ABC

AB BC

(Since, AB is the smallest side)

 

ACB BAC iii      

(Angle opposite to longer side is larger.)

Similarly,

In ,ADC

AD CD

(Since, CD is the largest side)

 

DCA DAC iv      

Adding (iii) and (iv) we get,

ACB DCA BAC DAC

BCD BAD

C A

A C

      

   

   

Hence, it is proved.

Question: 5

 

  

InFig  and bisects

Provethat

7.51, .

PR PQ PS QPR

PSR PSQ

Solution:

Given, PR>PQ and PS bisects

QPR

In ,PQR

PR>PQ (Given)

PQR PRQ      

(i) (



Angle opposite to the

longer side is larger.)

Now,

 

 bisects( )QPS RPS ii PS QPR      

,PQS

 

 PSR PQS QPS iii      

(Exterior angle of a

triangle is equal to the sum of opposite interior

angles.)

Similarly, in

,PSR

 

 PSQ PRS RPS iv       

(Exterior angle of a

triangle is equal to the sum of opposite interior angles)

Adding (i) and (ii) we get,

   

( rom andF

 

 )

PQR QPS PRQ RPS

Or PQS QPS PRS RPS

PSR PSQ iii iv

    

      

  

Hence, it is proved.

Question: 6

Show that of all line segments drawn from a given

point not on it, the perpendicular line segment is the

shortest.

Solution:

Let l be a line segment and B is a point lying on it.

We draw a line AB perpendicular to l. Let C (Different

from B) be a point on l.

In ,ABC

90B  

Now,



   

( )

180

90 90



A B C

A C B

C B

      

        

  



AB AC 

(Side opposite to the larger angle is

longer.)

Hence, it is proved.

Exercise 7.5 (4)

Question: 1

ABC is a triangle. Locate a point in the interior of

ABC

which is equidistant from all the vertices

ABC

Solution:

We know that circumcentre of a triangle is always

equidistant from all the vertices of that triangle.

So, circumcentre is the point where perpendicular

bisectors of all the sides of the triangle meet together.

Thus, in

ABC

, to find the circumcentre, we can draw

the perpendicular bisectors of sides AB, BC, and CA of

the triangle.

Hence, O is the point where these bisectors are

meeting together.

Therefore, O is the point which is equidistant from all

the vertices of

ABC

Question: 2

In a triangle locate a point in its interior which is

equidistant from all the sides of the triangle.

Solution:

We know that the point which is equidistant from all

the sides of a triangle is called the in centre of the

triangle.

So, in centre of a triangle is the intersection point of

the angle bisectors of the interior angles of the

triangle.

Let

ABC

be a triangle.

Then, in

ABC

, the in-centre of the triangle is the

angle bisectors of the interior angles of the triangle.

Let I is the point where these angle bisectors are

intersecting each other.

Therefore, I is the point which is equidistant from all

the sides of

ABC

Question: 3

In a huge park, people are concentrated at three points

(see Fig. 7.52):

A: where there are different slides and swings for

children,

B: near which a man-made lake is situated,

C: which is near to a large parking and exit

Where should an ice cream parlour be set up so that

maximum number of persons can approach

it?

(Hint: The parlour should be equidistant from A, B and

Solution:

To find the point where maximum number of persons

can approach the ice-cream parlour, we need to find

the point which is equidistant from A, B and C.

Now, A, B and C form a triangle. Thus circumcentre is

the only point that is equidistant from its vertices A, B

and C.

So, the ice-cream parlour should be set up at the

circumcentre O of

ABC

We know that circumcentre O of the triangle is the

intersecting point of perpendicular bisectors of the

sides of the triangle.

Question: 4

Complete the hexagonal and star shaped Rangolies

[see Fig. 7.53 (i) and (ii)] by filling them with as many

equilateral triangles of side 1 cm as you can. Count the

number of triangles in each case. Which has more

triangles?

Solution:

We can observe that hexagonal-shaped rangoli has 6

equilateral triangles in it.

   

side

3 3

4 4

 

Now, area of

OAB

 

3 25 3

25 cm

4 4

 

Thus, area of hexagonal-shaped rangoli

25 3 75 3

6 cm

4 2

  

Now, area of equilateral triangle having its side as

 

 

3 3

1 cm 1 cm

4 4

 

So, number of equilateral triangles of 1 cm side that

can be filled in this hexagonal-shaped rangoli

75 3

150

 

Therefore, star-shaped rangoli has 12 equilateral

triangles of side 5cm in it.

Here, area of star-shaped rangoli

 

12 5 75 3

   

So, number of equilateral triangles of 1cm side that

can be filled in star-shaped rangoli

75 3

300

 

Therefore, star-shaped rangoli has more equilateral

triangles in it.