Lesson: Triangles

Exercise 7.1

Question: 1

Which of the following is not a criterion for

congruence of triangles?

(a) SAS

(b) ASA

(c) SSA

(d) SSS

Solution:

c

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

If

,AB QR BC PR

and

CA PQ

, then:

(a)

ABC PQR

(b)

CBA PRQ

(c)

BAC RPQ

(d)

PQR BCA

Solution:

b

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

In

,ABC AB AC

and

50 .B

Then

C

is

equal to:

(a) 40°

(b) 50°

(c) 80°

(d) 130°

Solution:

b

Given: In

,ABC AB AC

and

50B

To find:

C

In

, .ABC AB AC

B C

(

Angles opposite to equal sides are equal).

Since

50B

(Given)

50C

Question: 4

In

,ABC BC AB

and

80 .B

Then

A

is

equal to:

(a) 80°

(b) 40°

(c) 50°

(d) 100°

Solution:

c

Given: In

,ABC BC AB

and

80 .B

To find:

A

In

,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

2

A

50A

Question: 5

In

,PQR R P

and

4 cmQR

and

5 cm.PR

Then the length of PQ is:

(a) 4 cm

(b) 5 cm

(c) 2 cm

(d) 2.5 cm

Solution:

a

Given:

, , 4 cmPOR R P QR

and

5 cm.PR

In

, .PQR R P

PQ QR

(

Sides opposite to equal angles of a

triangle are equal)

Given

4 cm

4 cm

PQ QR

PQ

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:

b

Given: In

,ABC AD

bisects

BAC

and D is a point

on side BC.

iBAD DAC

In

,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,

In

,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

°

4

DF F

DF E

DE E

DE D

Solution:

b

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

C

C

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

(c) 3.8 cm

(d) 3.4 cm

Solution:

d

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

In

,PQR

if

R Q

, then:

a

b

c

d

QR PR

PQ PR

PQ PR

QR PR

Solution:

b

In

, .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

(c) Congruent but not isosceles

(d) Neither congruent nor isosceles

Solution:

a

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:

a

b

c

d

BC EF

AC DE

AC EF

BC DE

Solution:

b

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

If

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

In

, 70PQR P

and

30 .R

Which side of this

triangle is the longest? Give reason for your answer.

Solution:

In

,PQR

70 , 30

180

70 30 180

180 70 30

180 100

80

P R

P Q R

Q

Q

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)

2

2

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

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

BD

and

CE

are its two medians. Show that

BD CE

.

Solution:

Given:

,AB AC BD

and CE are two medians.

In

ABD

and

,ACE

A A

Common

AB AC

Given

1 1

2 2

AB AC

AE AD

(

D

is midpoint of

,AC

and

E

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:

In

,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:

In

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:

In

,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

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:

In

ΔABC, AB AC

OB

and

OC

are the angle bisectors of

B

and

C

respectively.

In

,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

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:

In

,ABC AB AC

(Given)

ACB ABC

(Angles opposite to equal sides of

a triangle are equal.)

1 1

2 2

ACB ABC

2 1

(

BO

bisects

,ABC CO

bisects

ACB

)

In

,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

(

BO

bisects

,ABC CO

bisects

ACB

)

In

,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

In

CEB

and

CEA

B A

(As proved above)

( )

0

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:

In

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.

In

ADB

and

ADC

( )

0

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

(

BX

bisects

ABC

)

BA QP

(Given)

3 1 ... ii

(Alternate interior angles)

From (i) and (ii)

1 2

In

,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:

In

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

BC

at D. Prove that

2 .BC AD

Solution:

In right angle

,ABC

AB AC

Also,

BC

is the hypotenuse (

Hypotenuse longest

side)

90BAC

In

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

In

, 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

is

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

In

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

BE

are two Altitudes on sides

BC

and

AC

respectively. Prove that

.AE BD

Solution:

In triangles AEB and BDA,

1 2 90

In

,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:

In

,AOB

iAO OB AB

(Sum of two sides is greater

than the third side in a triangle)

In

,BOC

ii OB OC BC

(Sum of two sides is greater than

the third side in a triangle)

In

,COD

iiiOC OD CD

(Sum of two sides is greater than

the third side in a triangle)

In

,DOA

ivOD OA AD

(Sum of two sides is greater

than the third side in a triangle)

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

2

( )

2

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

In

,ABC

iAB BC AC

(Sum of two sides is greater

than the third side in a triangle)

In

,BCD

iiBC CD BD

(Sum of two sides is greater

than the third side in a triangle)

In

,CDA

iiiCD DA AC

(Sum of two sides is greater

than the third side in a triangle)

In

,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

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

1

.

2

BD AC

Show that

ABC

is a right angle.

Solution:

To prove:

90ABC

1

... i

2

BD AC

(Given)

and

1

..... ii

2

AD CD AC

(

D

is midpoint of

AC

)

From (i) and (ii)

AD BD CD

In

DAB

AD BD

1 A

(Angle opposite equal sides are

equal) ... …(iii)

In

Δ ,DBC

BD CD

2C

(Angle opposite equal sides are

equal) ... ….(iv)

In

,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

90

ABC

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

1

2

BE AC

.

Producing

BE

to

D

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

n

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,

in

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

In

DAC

and

DEC

3 90A

1 2

(Given)

DC DC

(Common)

DAC DEC

by AAS rule.

... iDA DE

(CPCT)

... iiCA CE

(CPCT)

In

BAC

AB AC

C B

(Angle opposite equal sides are equal)

Now,

180A B C

90 2 180

2 90

45

B B C

B

B

In

BED

5 180 4B

180 45 90

180 135

45

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

or

D B

.

Joining BD.

Now, in

,ABD

AD AB

1 3... i

(Angle opposite to the longer side is greater)

In

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

2

3

of a right angle.

Solution:

Let us consider

ABC

with BC as the longest side.

We need to prove

2

90

3

BAC

In

,ABC

BC AB

(As BC is the longest side)

..... iA C

(In a

Δ

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,

0

3

3 180

180

3

2

90

3

A A B C

A

A

2

3

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:

In

ABCD

,

AB AD

and

CB CD

We Join AC and BD.

Let AC and BD intersect at point O.

In

ABC

and

,ADC

AB AD

(Given)

BC CD

AC AC

ABC ADC

by SSS property

1 2

(CPCT)

In

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,

AC

is perpendicular bisector of

3 90

.B

BO DO