Lesson: Triangles

Exercise 7.1

Question: 1

Which of the following is not a criterion for

congruence of triangles?

(a) SAS

(b) ASA

(d) SSS

Solution:

The rule of congruence of triangles are SSS

(Side-Side-Side), SAS (Side-Angle-Side), ASA

(Angle-Side-Angle), AAS and RHS (Right

angle-Hypotenuse-Side). So, SSA is not a criterion for

congruence of triangles.

Question: 2

,AB QR BC PR 

and

CA PQ

, then:

(a)

ABC PQR  

(b)

CBA PRQ  

(c)

BAC RPQ  

(d)

PQR BCA  

Solution:

Since,

,AB QR BC RP 

and

CA PQ



A corresponds to Q, B corresponds to R &C

corresponds to P.

i.e.,

ABC QRP  

which can also be written as

.CBA PRQ  

Question: 3

,ABC AB AC 

and

50 .B  

Then

C

equal to:

(a) 40°

(b) 50°

(d) 130°

Solution:

Given: In

,ABC AB AC 

and

50B  

To find:

C

, .ABC AB AC 

B C   

(



Angles opposite to equal sides are equal).

Since

50B  

(Given)

   50C

Question: 4

,ABC BC AB 

and

80 .B  

Then

A

equal to:

(a) 80°

(b) 40°

(d) 100°

Solution:

Given: In

,ABC BC AB 

and

80 .B  

To find:

A

,ABC BC AB 

 

. iC A    

(∵ Angles opposite to equal sides

are equal)

Since, the sum of all the angles of a triangle is 180°

 

180

80 180 80

A B C

A C B

       

          

80 180A A       

(Using eqn. (i))

2 180 80 100A       

100



  

50A   

Question: 5

,PQR R P   

and

4 cmQR

and

5 cm.PR

Then the length of PQ is:

(a) 4 cm

(b) 5 cm

(d) 2.5 cm

Solution:

Given:

, , 4 cmPOR R P QR    

and

5 cm.PR

, .PQR R P   

PQ QR 

(



Sides opposite to equal angles of a

triangle are equal)

 

Given

4 cm

PQ QR

  

 

Hence, the length of PQ is 4 cm.

Question: 6

D is a point on the side BC of

ABC

such that AD

bisects

.BAC

Then,

(a)

BD CD

(b)

BA BD

(c)

BD BA

(d)

CD CA

Solution:

Given: In

,ABC AD

bisects

BAC

and D is a point

on side BC.

 

iBAD DAC    

,ACD ADB 

is an exterior angle.

 

iiADB DAC   

[



Exterior angle > interior opposites angle]

Using (i) & (ii), we have

ADB BAD  

BA BD 

(In a triangle, side opposite to the greater

angle is greater)

Similarly,

,ABD ADC 

is an exterior angle.

 

iiiADC BAD   

[



Exterior angle > interior

opposites angle]

Using (i) & (iii), we have

ADC CAD  

AC CD 

(In a triangle, side opposite to the greater

angle is greater).

Hence, using both the results we conclude that, option

(a), (c) & (d) cannot be true. Hence, the correct answer

is option (b).

Question: 7

It is given that

ABC FDE  

and

5 cm,AB 

40B  

and

80 .A  

Which one of the following

is true?

 

  

a 5 cm, 60°

b 5 cm, 60°

c 5 cm, 60°

d 5 cm, 0

DF F

DF E

DE E

DE D

Solution:

Given:

ABC FDE  

and

5 cmAB  ,

40B  

and

80A  

In ∆ABC,

80A  

and

40B  

Since, the sum of all the angles of a triangle is 180°

 

180

80 40 180

180 120

60 i

A B C

       

       

     

   

Since,

ABC FDE  

AB FD 

[By CPCT]

5 cm FD 

[Given

5 cmAB 

]

We can write

5 cmFD 

And also,

C E  

[By CPCT]

60 E   

[Using (i)]

Question: 8

Two sides of a triangle are of lengths 5 cm and 1.5 cm.

The length of the third side of the triangle cannot be:

(a) 3.6 cm

(b) 4.1 cm

(d) 3.4 cm

Solution:

Let us consider, ∆ABC with sides

5 cmBC 

and

1.5 cm.CA 

In a triangle, the difference of two sides < third side

and sum of two sides > third side.

BC CA AB  

and

BC CA AB 

5 1.5 AB  

and

5 1.5 AB 

3.5 AB 

and

6.5 AB

Here, all the options except (d) satisfy the inequality.

Question: 9

,PQR

R Q  

, then:

 

QR PR

PQ PR

QR PR





Solution:

, .PQR R Q   

PQ PR 

[Side opposite to greater angle is longer]

Question: 10

In triangles ABC and PQR,

,AB AC C P   

and

.B Q  

The two triangles are:

(a) Isosceles but not congruent

(b) Isosceles and congruent

(d) Neither congruent nor isosceles

Solution:

Given, In

, .ABC AB AC 

 

. i C B    

[Angles opposite to equal sides

are equal]

So, ∆ABC is an isosceles triangle.

Now, it is also given that

C P  

and

.B Q  

P Q  

[Using eq. (i)]

Now, In

,PQR

P Q  

[Proved above]

QR PR 

[Sides opposite to equal angles are equal]

So,

PQR

is an isosceles triangle.

Therefore, both triangles are isosceles.

But, corresponding sides of the triangles are not equal

and AAA is not a criterion for the congruence of

triangles. Hence, the triangles are not congruent.

Question: 11

In triangles ABC and

,DEF AB FD

and

.A D  

The two triangles will be congruent by SAS axiom if:

 

BC EF

AC DE

AC EF

BC DE



Solution:

Given, triangles

ABC

and

,DEF AB FD

and

.A D  

Two triangles are congruent by SAS axiom if two

corresponding sides and the included angle are equal.

i.e.,

ABC DEF  

Therefore, if

AC DE

, the triangles are congruent by

SAS axiom.

Exercise 7.2

Question: 1

In triangles ABC and

,PQR A Q  

and

.B R  

Which side of ∆PQR should be equal to side AB of

∆ABC so that the two triangles are congruent?

Give reason for your answer.

Solution:

Given,

∆ABC & ∆PQR,

A Q  

and

.B R  



AB and QR are the included sides.

So, if

AB QR

, then

ABC QRP  

(By ASA axiom).

Question: 2

In triangles ABC and PQR,

A Q  

and

.B R  

Which side of ∆PQR should be equal to side BC of

∆ABC so that the two triangles are congruent?

Give reason for your answer.

Solution:

Given, in triangles ABC and PQR,

A Q  

and

.B R  



RP is corresponding side to BC.

So, if

BC RP

, then

ABC QRP  

(By AAS axiom).

Question: 3

“If two sides and an angle of one triangle are equal to

two sides and an angle of another triangle, then the

two triangles must be congruent.” Is the statement true?

Why?

Solution:

No; because with respect to congruency rule, two

sides and the included angle of one triangle are equal

to two sides and the included angle of another triangle,

which is called SAS rule.

Question: 4

“If two angles and a side of one triangle are equal to

two angles and a side of another triangle, then the two

triangles must be congruent.” Is the statement true?

Why?

Solution:

No, because side of one triangle must be equal to the

corresponding side of the other triangle.

Question: 5

Is it possible to construct a triangle with sides

measuring 4 cm, 3 cm and 7 cm? Give reason for your

answer.

Solution:

No; because in a triangle the sum of any two sides is

greater than the third side, but

4 3 7.

Question: 6

It is given that

.ABC RPQ  

Is it true to say that

BC QR

? Why?

Solution:

No; because as per the congruency rule, sides &

angles of one triangle should be equal to the

corresponding sides and angle of the other triangle.

Here,

ABC RPQ  

, , AB RP BC PQ AC RQ   

[By CPCT]

Hence, BC is not necessarily equal to QR.

Question: 7

PQR EDF  

, then is it true to say that

PR EF

Give reason for your answer.

Solution:

Yes; because as per the congruency rule, sides &

angles of one triangle should be equal to the

corresponding sides and angle of the other triangle. If

,PQR EDF  

, , PQ ED QR DF RP FE   

[By CPCT]

Question: 8

, 70PQR P   

and

30 .R  

Which side of this

triangle is the longest? Give reason for your answer.

Solution:

,PQR

 

70 , 30

180

70 30 180

180 70 30

180 100

P R

P Q R

     

      

       

       

   

 



Therefore Q is largest angle.

This means PR is the longest. (Side opposite to the

largest angle is longest)

Question: 9

AD is a median of the triangle ABC. Is it true that

2AB BC CA AD  

? Give reason for your answer.

Solution:

In triangle ABD,

 

... 1AB BD AD 

(Sum of any two sides of a triangle is greater than the

third side)

In triangle ACD,

 

... 2AC CD AD 

(Sum of any two sides of a triangle is greater than the

third side)

Adding eq. (1) and (2)

 

AB BD AC CD AD

AB BD CD AC AD

AB BC AC AD BD CD BC

   

    

     

Hence, it is proved.

Question: 10

M is a point on side BC of a triangle ABC such that

AM is the bisector of

.BAC

Is it true to say that

perimeter of the triangle is greater than 2AM? Give

reason for your answer.

Solution:

In triangle ABC, AM is the bisector of angle BAC.

In triangle ABM. we have

 

.. 1AB BM AM  

[Sum of any two sides of a

triangle is always greater than its third side]

Similarly, in triangle AMC, we have

AC CM AM 

[Sum of any two sides of a triangle

is always greater than its third side] … (2)

On adding (1) and (2) we get,

( ) 2

AB BM AC CM AM AM

AB BM CM AC AM

AB BC AC AM

    

   

  

Hence, it is proved

Question: 11

Is it possible to construct a triangle with sides

measuring 9 cm, 7 cm and 17 cm? Give reason for

your answer.

Solution:

No. Because, in any triangle, the sum of any two sides

should be greater than the third side.

But in this case,

9 7 16 17.  

Question: 12

Is it possible to construct a triangle with sides

measuring8 cm, 7 cm and4 cm? Give reason for your

answer.

Solution:

Yes, because in each case the sum of two sides is

greater than the third side.

. ., 7 4 8,

8 4 7,

7 8 4.

i e  

 

Exercise 7.3

Question: 1

ABC is an isosceles triangle with

AB AC

and

are its two medians. Show that

BD CE

Solution:

Given:

 ,AB AC BD

and CE are two medians.

ABD

and

,ACE

A A  

Common

AB AC

Given

1 1

2 2

AB AC 

AE AD 

(

D

is midpoint of

,AC

and

is midpoint of

.AB

)

ABD ACE   

(By SAS property)

BD CE

(By CPCT)

Question: 2

In Fig.7.4, D and E are points on side BC of a ∆ABC

such that

BD CE

and

.AD AE

Show that

.ABD ACE  

Solution:

Given: In ∆ABC,

BD EBC

and

AD AE

In ∆ADE

AD AE

(Given)

2 1  

(Angles opposite to equal sides of a triangle

are equal.)

1 3 180 ................(i)

2 4 180 ...............(ii)

    

[Linear pair axiom]

From (i) and (ii)

1 3 2 4

1 2

3 4

      

  

 

 

   



In ∆ABD and ∆ACE

AD AE

(Given)

3 4  

(As proved above)

BD CE

(Given)

Δ ΔABD ACE 

(By

SAS

property)

Question: 3

CDE is an equilateral triangle formed on a side CD of

a square ABCD (Fig. 7.5). Show that

.ADE BCE  

Solution:

,DEC DE CE 

(All sides of equilateral ∆ are

equal)

...(i)4 3   

[Angles opposite to equal sides of a

triangle are equal.]

And,

2 1 90    

… (ii)

From (i) and (ii),

4 2 1 3      

BCE ADE   

AD BC

(Sides of square)

Δ Δ ADE BCE 

by SAS property.

Question 4

In Fig. 7.6,

,BA AC DE DF 

such that

BA DE

and

.BF EC

Show that

.ABC DEF  

Solution:

ABC

and

DEF

90A D    

(Given)

BF CE 

(Sides opposite to equal angles are equal)

BF CF CE CF   

(Adding CF both the sides)

BC EF 

AB DE

(Given)

ABC DEF   

(By RHS property)

Question: 5

Q is a point on the side SR of a ∆PSR such

that

PQ PR

. Prove that

.PS PQ

Solution:

,PRQ

PR PQ

1 R   

(Angles opposite to equal sides of a

triangle are equal.)

But,

1 S  

(

1

is an exterior angle for

PQS

)

( 1 )R S R     

PS PR 

(



In any triangle, side opposite to larger

angle is longer)

)( PR PQPS PQ  

Hence, the result.

Question: 6

S is any point on side QR of ∆PQR. Show that

2 .PQ QR RP PS  

Solution:

In ∆PQS,

 

... iPQ QS PS 

(



Sum of two sides of a ∆ is

greater than the third side)

In ∆PSR,

i( i)RS PR PS  

(



Sum of two sides of a

∆ is greater than the third side)

Adding eq. (i) and eq. (ii), we get

2 ( ).

PQ QS SR PR PS

PQ QR PR PS QS SR QR

   

     

Question: 7

D is any point on side AC of a ∆ABC with

.AB AC

Show that

.CD BD

Solution:

In ∆ABC, Join BD

AB AC

(Given)

C ABC   

(Angles opposite to equal sides of a

triangle are equal.)

But,

) (CBD ABCAB A CBDC BD       



C CBD  

(

C ABC  

, as proved above)

Then, in ∆DBC,

BD CD

(As the side opposite to greater angle is longer.)

Question: 8

In Fig. 7.7, l ||m and M is the midpoint of a line

segment AB. Show that M is also the midpoint of any

line segment CD, having its end points on l and m,

respectively.

Solution:

Given: l



m, M is midpoint of line segment AB, i.e.,

AM MB

We need to prove that

.MC MD

Let us consider, ∆AMC and ∆BMD.

Since,

l m

and AB is the transversal,

1 2  

(Alternate interior angles)

AM BM

(Given)

3 4  

(Vertical opposite angles)

AMC BMD  Δ

by ASA property

MC MD

(By CPCT)

Question 9

Bisectors of the angles B and C of an isosceles triangle

with

AB AC

intersect each other at O. BO is

produced to a point M. Prove that

.MOC ABC  

Solution:

ΔABC, AB AC

and

are the angle bisectors of

B

and

C

respectively.

,ABC

AB AC

(Given)

ACB ABC   

(Angles opposite to equal sides of

a triangle are equal.)

1 1

2 2

ACB ABC   

 

2 1... i   

bisects

BO ABC

CO ACB



 

 



 

1 2MOC    

(Exterior angle property of

triangle.)

2 1MOC   

(Using (i))

( 2 1)MOC ABC ABC      

Hence, it is proved.

Question 10

Bisectors of the angles B and C of an isosceles triangle

ABC, with

,AB AC

intersect each other at O. Show

that the external angle adjacent to

ABC

is equal to

BOC

Solution:

,ABC AB AC 

(Given)

ACB ABC  

(Angles opposite to equal sides of

a triangle are equal.)

1 1

2 2

ACB ABC   

2 1   

(

bisects

,ABC CO

bisects

ACB

)

,BOC

1 2 180BOC      

(Angle sum property of

triangle.)

1 1 180BOC       

(

1 2,  

as proved

above)

2 1 180BOC     

...180 (i)ABC BOC     

(

BO

bisects

ABC

)

But,

....180 (ii)ABC DBA    

(



DBC is a line segment)

On comparing eqn. (i) and (ii), we get,

DBA BOC   

Question 11

In Fig. 7.8, AD is the bisector of

BAC

. Prove

that

.AB BD

Solution:

In ∆ABC, AD is angle bisector of

.BAC

Since AD is bisector of

BAC

1 2  

But,

2 ADB  

[

ADB

is an exterior angle to

∆ADC]

1 ( 1 )2ADB    

∴ In ∆ADB,

AB BD

(Side opposite to the larger angle is longer)

Hence, it is proved.

Exercise 7.4

Question: 1

Find all the angles of an equilateral triangle.

Solution:

Given,

,ABC AB AC 

is an equilateral triangle.

ACB ABC  

(Angle opposite to equal sides of a

triangle is equal.)

1 1

2 2

ACB ABC   

2 1   

(

bisects

,ABC CO

bisects

ACB

)

,BOC

1 2 180BOC      

(Angle sum property of a

triangle)

1 1 180BOC       

(

1 2,  

as proved

above)

2 1 180BOC     

...180 (i)ABC BOC     

(

BO

bisects

ABC

)

But,

....180 (ii)ABC DBA    

(

DBC

is a line segment)

On comparing equations (i) and (ii), we get,

DBA BOC   

Question: 2

The image of an object placed at a point A before a

plane mirror LM is seen at the point B by an observer

at D as shown in Fig. 7.12. Prove that the image is as

far behind the mirror as the object is in front of the

mirror.

[Hint: CN is perpendicular to the mirror.

Also, angle of incidence



angle of reflection].

Solution:

To prove:

AE BE

Proof:

&CN LM AB LM 

AB NC 

 

... iA i  

(Alternate interior angles)

 

... iiB r  

(Corresponding angles)

 

... iiii r  

(Incident angle



reflected angle)

Using (i), (ii) and (iii), we get,

A B  

CEB

and

CEA

B A  

(As proved above)

( )

1 2 90 AB LM    

CE CE

(Common)

CEB CEA  

by AAS rule

BE AE 

(By CPCT)

Question: 3

ABC is an isosceles triangle with

AB AC

and D is

a point on BC such that

AD BC

(Fig. 7.13). To

prove that

,BAD CAD  

a student proceeded as

follows:

ABD

and

,ACD

AB AC

(Given)

B C  

(Because

AB AC

) and

ADB ADC  

Therefore,

ABD ACD  

(AAS)

So,

BAD CAD  

(CPCT)

What is the error in the above proof?

(Hint: Recall how

B C  

is proved when

AB AC

Solution:

AB AC

cannot conclude that

ABD ACD  

Here is the correct proof.

ADB

and

ADC

( )

1 2 90 AD BC    

AB AC

(Given)

AD AD

(Common)

ADB ADC  

(By RHS)

BAD CAD  

(By CPCT)

Question: 4

P is a point on the bisector of

ABC

If a line through P, parallel to BA, meet BC at Q,

prove that BPQ is an isosceles triangle.

Solution:

 

2 3 ... i  

(

bisects

ABC

)

BA QP

(Given)

 

3 1 ... ii  

(Alternate interior angles)

From (i) and (ii)

1 2  



,BPQ

BQ PQ

(Converse of isosceles



property)

BPQ

is an isosceles triangle.

Question: 5

ABCD is a quadrilateral in which

AB BC

and

.AD CD

Show that BD bisects both the angles ABC

and ADC.

Solution:

DAB

and

.DCB

AB CB

(Given)

AD CD

(Given)

DB DB

(Common)





DAB 



DCB

by SSS property

1 2

3 4

  

DB

bisects

D

and

B

Question: 6

ABC is a right triangle with

.AB AC

Bisector of

A

meets

at D. Prove that

2 .BC AD

Solution:

In right angle

,ABC

AB AC

Also,

is the hypotenuse (



Hypotenuse longest

side)



90BAC  



CAD

and

BAD

AC AB

(Given)

1 2  

(

AD

is an angle bisector of

A

)

AD AD

(Common)





CAD  

BAD

by SAS property

CD BD 

(By CPCT)

and

....(i)ACD ABD  

(By CPCT)

In ∆ABC,

180A B C      

90 180B B       

(

90A  

and using (i))

2 180 90 90B       

45 45B C       

(Using (i))

Also,

1 45  

and

2 45  

(

AD

bisects

A

and

90A  

)

1 C   

and

2 B  



, ACD AD CD 

Similarly, In

,ABD AD BD 

(Sides opposite to

equal angles are equal)

Hence,

2BC BD CD AD AD AD    

(Using

above equations)

Therefore,

2 .BC AD

Question: 7

O is a point in the interior of a square ABCD such that

OAB is an equilateral triangle. Show that

OCD

an isosceles triangle.

Solution:

 

1 2 90 ..... i    

(

ABCD

is a square)

 

3 4 60 ...... ii    

(

OAB

is an equilateral

triangle)

Using (i) and (ii)

1 3 2 4 90 60          

5 6 30     

DAO

and

CBO

AD BC

(Given)

5 6  

(As proved above)

AO BO

(Given)

DAO CB  

by SAS property

OD OC

OCD 

is an isosceles triangle.

Question: 8

ABC and DBC are two triangles on the same base BC

such that A and D lie on the opposite sides of BC,

AB=AC and DB=DC. Show that AD is the

perpendicular bisector of BC.

Solution:

In triangle ABD and ACD

AB AC

(Given)

DB DC

(Given)

AD DA

(Common)

Therefore,

ABD

is congruent to

.ACD

(By SSS

congruency rule)

So,

 

By CPCT1 2  

Now, in triangle AOB and AOC

AB = AC (Given)

 

Proved above1 2  

AO AO

(Common side)

Therefore,

AOB

is congruent to

AOC

So,

 

By CPCT3 4  

and

BO OC

(By CPCT)

Since

3 4  

Therefore,

3 4 90    

Hence, proved that AD is the perpendicular bisector of

BC.

Question: 9

ABC

is an isosceles triangle in which

,AC BC AD

and

are two Altitudes on sides

and

respectively. Prove that

.AE BD

Solution:

In triangles AEB and BDA,

1 2 90    

,ABC

AC BC

[Given]

4 3   

[By isosceles triangle property]

AB BA

(Common)

AEB BDA  

(By AAS rule)

AE BD 

(By CPCT)

Question: 10

Prove that sum of any two sides of a triangle is greater

than twice the median.

Solution:

Let us consider a ∆ABC such that AD is a median on

BC i.e.

.BD BC

We need to prove that

2AB AC AD 

Let’s produce AD to E such that

,DE AD

and join EC.

Now, in

ADB

and

EDC

AD ED

(By construction)

1 2  

(Vertically opposite angle)

DB DC

(Given)

ADB EDC  

By SAS property

So,

AB EC

(By CPCT)

Now, In

,AEC

AC CE AE

AC CE AD DE

 

   

2AC AB AD  

(

AB CE

as proved above

and

AD DE

by construction)

Hence, it is proved.

Question: 11

Show that in quadrilateral

 

, 2 .ABCD AB BC CD DA BD AC    

Solution:

,AOB

 

iAO OB AB  

(Sum of two sides is greater

than the third side in a triangle)

,BOC

 

ii OB OC BC  

(Sum of two sides is greater than

the third side in a triangle)

,COD

 

iiiOC OD CD  

(Sum of two sides is greater than

the third side in a triangle)

,DOA

 

ivOD OA AD  

(Sum of two sides is greater

than the third side in a triangle)

Adding (i), (ii), (iii) and (iv)

 

        

  

       

     

( )

AO OB OB OC OC OD OD OA

AB BC CD DA

AO OC BO OD AB BC CD DA

AC BD AB BC CD DA

Hence, it is proved.

Question: 12

Show that in a quadrilateral ABCD,

AB+BC+CD+DA>AC+BD.

Solution:

To Prove: In quadrilateral ABCD,

AB BC CD DA AC BD    

,ABC

 

iAB BC AC  

(Sum of two sides is greater

than the third side in a triangle)

,BCD

 

iiBC CD BD  

(Sum of two sides is greater

than the third side in a triangle)

,CDA

 

iiiCD DA AC  

(Sum of two sides is greater

than the third side in a triangle)

,DAB

 

ivAD AB BD  

(Sum of two sides is greater

than the third side in a triangle)

Adding (i), (ii), (iii) and (iv)

( )

( ) ( )2 2

AB BC BC CD CD DA AD AB

AC BD AC BD

AB BC CD DA AC BD

       

  

    



Hence, it is proved.

Question: 13

In a triangle ABC, D is the mid-point of side AC such

that

BD AC

Show that

ABC

is a right angle.

Solution:

To prove:

90ABC  

 

... i

BD AC

(Given)

and

 

..... ii

AD CD AC 

(

D

is midpoint of

)

From (i) and (ii)

AD BD CD 

DAB

AD BD

1 A   

(Angle opposite equal sides are

equal) ... …(iii)

Δ ,DBC

BD CD

2C  

(Angle opposite equal sides are

equal) ... ….(iv)

,ABC

180A ABC C      

(Angle sum property of

triangle)

1 2 180ABC       

 

1, 2A C     

1 2 180ABC       

180ABC ABC     

( 1 2 )ABC    

2 180

ABC

   

Hence, it is proved.

Question: 14

In a right triangle, prove that the line-segment joining

the mid-point of the hypotenuse to the opposite vertex

is half the hypotenuse.

Solution:

Let us consider a right

ABC

with

90ABC  

and let E be the midpoint of hypotenuse AC.

We need to prove

BE AC

Producing

s.t.

ED BE

Join A to D and C to D

Since,

=ED EB

(By construction)

and

EA EC

(Given)

ABCD 

is a parallelogram

90ABC  

Also, in a parallelogram, if one angle is right, then all

angle are right.



Parallelogram

ABCD

rectangle

AC BD

(Diagonals of a rectangle are equal)

1 1

2 2

AC BD 

1 1

2 2

AC BE BE ED BD

 

   

 

 



Hence, it is proved.

Question: 15

Two lines l and m intersect at the point O and P is a

point on a line n passing through the point O such that

P is equidistant from l and m. Prove that n is the

bisector of the angle formed by l and m.

Solution:

We draw

PQ l

and

.PR m

Now, in

OQP

and

,ORP

1 2 90    

(By construction)

OP OP

(Common)

PQ PR

(Given P is equidistant from l and m)

OQP ORP  

By RHS property

By CPCT

3 4  



is the bisector of

QOR

Question: 16

Line segment joining the mid-points M and N of

parallel sides AB and DC, respectively of a trapezium

ABCD is perpendicular to both the sides AB and DC.

Prove that

AD BC

Solution:

Joining AN and BN,

AMN

and

,BMN

AM BM

(Given)

1 2 90    

NM NM

(Common)

AMN BMN  

by SAS property

3 4   

and,

...(i)AN BN

(By CPCT)

Now,

90DNM CNM    

5 3 6 4       

....... (5 6 (ii) 3 4)      

Now, In

ADN

and

,BCN

DN CN

(Given)

5 6  

(Using (ii))

AN BN

(Using (i))

ADN BCN  

by SAS property

AD BC 

(By CPCT)

Question: 17

ABCD is a quadrilateral such that diagonal AC bisects

the angles A and C. Prove that

AB AD

and

.CB CD

Solution:

Given: In quadrilateral ABCD, diagonal AC bisects

A

and

.C

i.e.,

1 2  

and

3 4  

Now, in

ABC

and

,ADC

1 2  

(Given)

AC AC

3 4  

(Given)

ABC ADC 

by ASA Property

AB AD

(By CPCT)

CB CD

(By CPCT)

Question: 18

If ABC is a right angled triangle such that

AB AC

and bisector of angle C intersects the side AB at D.

Prove that

.AC AD BC 

Solution:

Given: In figure ABC is a right angle,

AB AC

, CD

is bisector of

C

Construction: Draw

DE BC

In right angle

,ABC

AB AC



BC is the hypotenuse

90A   

DAC

and

DEC

3 90A    

1 2  

(Given)

DC DC

(Common)

DAC DEC  

by AAS rule.

 

... iDA DE 

(CPCT)

 

... iiCA CE

(CPCT)

BAC

AB AC

C B   

(Angle opposite equal sides are equal)

Now,

180A B C      

 

90 2 180

2 90

B B C

       

   



BED

 

5 180 4B      

 

180 45 90

180 135

    

   

 

5B  

 

... iiiDE BE 

(Sides opposite equal angles are equal)

From (i) and (iii)

 

... ivDA DE BE 

Now,

BC CE BE 

CA DA 

(Using (ii) and (iii)

BC AD AC  

Hence, it is proved.

Question: 19

AB and CD are the smallest and largest sides of a

quadrilateral ABCD. Out of

B

and

D

decide

which is greater.

Solution:

AB is the shortest and CD is the longest side of ABCD.

We need to prove

B D  

D B  

Joining BD.

Now, in

,ABD

AD AB

 

1 3... i   

(Angle opposite to the longer side is greater)

BCD

CD BC

 

2 4...... ii   

 

2 4...... ii   

(Angle opposite to the longer side is greater)

Adding equation (i) and (ii), we get

1 2 3 4

B D

      

  

Question: 20

Prove that in a triangle, other than in an equilateral

triangle, angle opposite the longest side is greater than

of a right angle.

Solution:

Let us consider

ABC

with BC as the longest side.

We need to prove

BAC   

,ABC

BC AB

(As BC is the longest side)

 

..... iA C   

(In a

angle opposite the longer side is greater)

Similarly,

BC AC

(As BC is the longest side)

 

... iiBA  

Adding (i) and (ii) we get,

2 A B C    

Adding

A

on both the sides,

3 180

180

A A B C

      

 

  

 

A  

right angle.

Question: 21

ABCD is a quadrilateral such that AB=AD and CB=CD.

Prove that AC is the perpendicular bisector of BD.

Solution:

ABCD

AB AD

and

CB CD

We Join AC and BD.

Let AC and BD intersect at point O.

ABC

and

,ADC

AB AD

(Given)

BC CD

AC AC



ABC ADC  

by SSS property

1 2  

(CPCT)

AOB

and

AOD

AB AD

(Given)

1 2  

(As proved above)

AO AO

AOB AOD  

by SAS property

BO DO

(CPCT)

3 4  

But,

3 4 180    

(Linear pair axiom)

2 3 180

3 90

4 90

  

   

Hence,

is perpendicular bisector of

3 90

BO DO

  

 

 



 