# NCERT Solutions for Class 9 Maths Chapter 7 Triangles| PDF Download

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**Chapter 7 Triangles Class 9 Maths NCERT Solutions**that is going to help you in understanding the concepts in a better way and prepare for the examinations. You can download PDF of NCERT Solutions for Class 9 Maths Chapter 7 Triangles in order to learn at your ease. You can also complete your homework on time through the help of these NCERT Solutions and able to solve the difficult problems in a given in a exercise.**Class 9 Maths NCERT Solutions**will help in developing your problem solving skills and be aware of the concepts. These solutions are prerequisites before solving exemplar problems and going for supplementary Maths Books.

#### Page No: 118

**Exercise 7.1**

1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that Î”ABC ≅ Î”ABD. What can you say about BC and BD?

**Answer**

Given,

AC = AD and AB bisects ∠A

To prove,

Î”ABC ≅ Î”ABD

Proof,

In Î”ABC and Î”ABD,

AB = AB (Common)

AC = AD (Given)

∠CAB = ∠DAB (AB is bisector)

Therefore, Î”ABC ≅ Î”ABD by SAS congruence condition.

BC and BD are of equal length.

Page No: 119

2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Fig. 7.17). Prove that

(i) Î”ABD ≅ Î”BAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

**Answer**

Given,

AD = BC and ∠DAB = ∠CBA

(i) In Î”ABD and Î”BAC,

AB = BA (Common)

∠DAB = ∠CBA (Given)

AD = BC (Given)

Therefore, Î”ABD ≅ Î”BAC by SAS congruence condition.

(ii) Since, Î”ABD ≅ Î”BAC

Therefore BD = AC by CPCT

(iii) Since, Î”ABD ≅ Î”BAC

Therefore ∠ABD = ∠BAC by CPCT

3. AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.

**Answer**

Given,

AD and BC are equal perpendiculars to AB.

To prove,

CD bisects AB

Proof,

In Î”AOD and Î”BOC,

∠A = ∠B (Perpendicular)

∠AOD = ∠BOC (Vertically opposite angles)

AD = BC (Given)

Therefore, Î”AOD ≅ Î”BOC by AAS congruence condition.

Now,

AO = OB (CPCT). CD bisects AB.

4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that Î”ABC ≅ Î”CDA.

**Answer**

Given,

l || m and p || q

To prove,

Î”ABC ≅ Î”CDA

Proof,

In Î”ABC and Î”CDA,

∠BCA = ∠DAC (Alternate interior angles)

AC = CA (Common)

∠BAC = ∠DCA (Alternate interior angles)

Therefore, Î”ABC ≅ Î”CDA by ASA congruence condition.

5. Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see Fig. 7.20). Show that:

(i) Î”APB ≅ Î”AQB

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

Given,

l is the bisector of an angle ∠A.

BP and BQ are perpendiculars.

(i) In Î”APB and Î”AQB,

∠P = ∠Q (Right angles)

∠BAP = ∠BAQ (l is bisector)

AB = AB (Common)

Therefore, Î”APB ≅ Î”AQB by AAS congruence condition.

(ii) BP = BQ by CPCT. Therefore, B is equidistant from the arms of ∠A.

Page No: 120

6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

**Answer**

Given,

AC = AE, AB = AD and ∠BAD = ∠EAC

To show,

BC = DE

Proof,

∠BAD = ∠EAC (Adding ∠DAC both sides)

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠EAD

In Î”ABC and Î”ADE,

AC = AE (Given)

∠BAC = ∠EAD

AB = AD (Given)

Therefore, Î”ABC ≅ Î”ADE by SAS congruence condition.

BC = DE by CPCT.

7. AB is a line segment and P is its mid-point. 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) Î”DAP ≅ Î”EBP

(ii) AD = BE

**Answer**

Given,

P is mid-point of AB.

∠BAD = ∠ABE and ∠EPA = ∠DPB

(i) ∠EPA = ∠DPB (Adding ∠DPE both sides)

∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒ ∠DPA = ∠EPB

In Î”DAP ≅ Î”EBP,

∠DPA = ∠EPB

AP = BP (P is mid-point of AB)

∠BAD = ∠ABE (Given)

Therefore, Î”DAP ≅ Î”EBP by ASA congruence condition.

(ii) AD = BE by CPCT.

8. In right triangle ABC, right angled at C, M is the mid-point 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 (see Fig. 7.23). Show that:

(i) Î”AMC ≅ Î”BMD

(ii) ∠DBC is a right angle.

(iii) Î”DBC ≅ Î”ACB

(iv) CM = 1/2 AB

**Answer**

Given,

∠C = 90°, M is the mid-point of AB and DM = CM

(i) In Î”AMC and Î”BMD,

AM = BM (M is the mid-point)

∠CMA = ∠DMB (Vertically opposite angles)

CM = DM (Given)

Therefore, Î”AMC ≅ Î”BMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)

Therefore, AC || BD as alternate interior angles are equal.

Now,

∠ACB

⇒ 90° + ∠B = 180°

⇒ ∠DBC = 90°

(iii) In Î”DBC and Î”ACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC (byy CPCT, already proved)

Therefore, Î”DBC ≅ Î”ACB by SAS congruence condition.

(iv) DC = AB (Î”DBC ≅ Î”ACB)

⇒ DM = CM = AM = BM (M is mid-point)

⇒ DM + CM = AM + BM

⇒ CM + CM = AB

⇒ CM = 1/2AB

Page No: 123

**Exercise 7.2**

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

**Answer**

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

⇒ 1/2∠B = 1/2∠C

⇒ ∠OBC = ∠OCB (Angle bisectors.)

⇒ OB = OC (Side opposite to the equal angles are equal.)

(ii) In Î”AOB and Î”AOC,

AB = AC (Given)

AO = AO (Common)

OB = OC (Proved above)

Therefore, Î”AOB ≅ Î”AOC by SSS congruence condition.

∠BAO = ∠CAO (by CPCT)

Thus, AO bisects ∠A.

2. In Î”ABC, AD is the perpendicular bisector of BC (see Fig. 7.30). Show that Î”ABC is an isosceles triangle in which AB = AC.

**Answer**

Given,

AD is the perpendicular bisector of BC

To show,

AB = AC

Proof,

In Î”ADB and Î”ADC,

AD = AD (Common)

∠ADB = ∠ADC

BD = CD (AD is the perpendicular bisector)

Therefore, Î”ADB ≅ Î”ADC by SAS congruence condition.

AB = AC (by CPCT)

Page No: 124

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.

**Answer**

Given,

BE and CF are altitudes.

AC = AB

To show,

BE = CF

Proof,

In Î”AEB and Î”AFC,

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

AB = AC (Given)

Therefore, Î”AEB ≅ Î”AFC by AAS congruence condition.

Thus, BE = CF by CPCT.

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) Î”ABE ≅ Î”ACF

(ii) AB = AC, i.e., ABC is an isosceles triangle.

**Answer**

Given,

BE = CF

(i) In Î”ABE and Î”ACF,

∠A = ∠A (Common)

∠AEB = ∠AFC (Right angles)

BE = CF (Given)

Therefore, Î”ABE ≅ Î”ACF by AAS congruence condition.

(ii) Thus, AB = AC by CPCT and therefore ABC is an isosceles triangle.

5. ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.

**Answer**

Given,

ABC and DBC are two isosceles triangles.

To show,

∠ABD = ∠ACD

Proof,

In Î”ABD and Î”ACD,

AD = AD (Common)

AB = AC (ABC is an isosceles triangle.)

BD = CD (BCD is an isosceles triangle.)

Therefore, Î”ABD ≅ Î”ACD by SSS congruence condition. Thus, ∠ABD = ∠ACD by CPCT.

6. Î”ABC 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.

**Answer**

Given,

AB = AC and AD = AB

To show,

∠BCD is a right angle.

Proof,

In Î”ABC,

AB = AC (Given)

⇒ ∠ACB = ∠ABC (Angles opposite to the equal sides are equal.)

In Î”ACD,

AD = AB

⇒ ∠ADC = ∠ACD (Angles opposite to the equal sides are equal.)

Now,

In Î”ABC,

∠CAB + ∠ACB

⇒ ∠CAB + 2∠ACB = 180°

⇒ ∠CAB = 180° - 2∠ACB --- (i)

Similarly in Î”ADC,

∠CAD = 180° - 2∠ACD --- (ii)

also,

∠CAB + ∠CAD = 180° (BD is a straight line.)

Adding (i) and (ii)

∠CAB + ∠CAD = 180° - 2∠ACB + 180° - 2∠ACD

⇒ 180° = 360° - 2∠ACB - 2∠ACD

⇒ 2(∠ACB + ∠ACD) = 180°

⇒ ∠BCD = 90°

7. ABC is a right angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

**Answer**

Given,

∠A = 90° and AB = AC

A/q,

AB = AC

⇒ ∠B = ∠C (Angles opposite to the equal sides are equal.)

Now,

∠A + ∠B + ∠C = 180° (Sum of the interior angles of the triangle.)

⇒ 90° + 2∠B = 180°

⇒ 2∠B = 90°

⇒ ∠B = 45°

Thus, ∠B = ∠C = 45°

8. Show that the angles of an equilateral triangle are 60° each.

**Answer**

BC = AC = AB (Length of all sides is same)

⇒ ∠A = ∠B = ∠C (Sides opposite to the equal angles are equal.)

Also,

∠A + ∠B + ∠C = 180°

⇒ 3∠A = 180°

⇒ ∠A = 60°

Therefore, ∠A = ∠B = ∠C = 60°

Thus, the angles of an equilateral triangle are 60° each.

Page No: 128

**Exercise 7.3**

1. Î”ABC and Î”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 intersect BC at P, show that

(i) Î”ABD ≅ Î”ACD

(ii) Î”ABP ≅ Î”ACP

(iii) AP bisects ∠A as well as ∠D.

(iv) AP is the perpendicular bisector of BC.

**Answer**

Given,

Î”ABC and Î”DBC are two isosceles triangles.

(i) In Î”ABD and Î”ACD,

AD = AD (Common)

AB = AC (Î”ABC is isosceles)

BD = CD (Î”DBC is isosceles)

Therefore, Î”ABD ≅ Î”ACD by SSS congruence condition.

(ii) In Î”ABP and Î”ACP,

AP = AP (Common)

∠PAB = ∠PAC (Î”ABD ≅ Î”ACD so by CPCT)

AB = AC (Î”ABC is isosceles)

Therefore, Î”ABP ≅ Î”ACP by SAS congruence condition.

(iii) ∠PAB = ∠PAC by CPCT as Î”ABD ≅ Î”ACD.

AP bisects ∠A. --- (i)

also,

In Î”BPD and Î”CPD,

PD = PD (Common)

BD = CD (Î”DBC is isosceles.)

BP = CP (Î”ABP ≅ Î”ACP so by CPCT.)

Therefore, Î”BPD ≅ Î”CPD by SSS congruence condition.

Thus, ∠BDP = ∠CDP by CPCT. --- (ii)

By (i) and (ii) we can say that AP bisects ∠A as well as ∠D.

(iv) ∠BPD = ∠CPD (by CPCT as Î”BPD ≅ Î”CPD)

and BP = CP --- (i)

also,

∠BPD + ∠CPD = 180° (BC is a straight line.)

⇒ 2∠BPD = 180°

⇒ ∠BPD = 90° ---(ii)

From (i) and (ii),

AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC (ii) AD bisects ∠A.

**Answer**

Given,

AD is an altitude and AB = AC

(i) In Î”ABD and Î”ACD,

∠ADB = ∠ADC = 90°AB = AC (Given)

AD = AD (Common)

Therefore, Î”ABD ≅ Î”ACD by RHS congruence condition.

Now,

BD = CD (by CPCT)

Thus, AD bisects BC

(ii) ∠BAD = ∠CAD (by CPCT)

Thus, AD bisects ∠A.

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 Î”PQR (see Fig. 7.40). Show that:

(i) Î”ABM ≅ Î”PQN

(ii) Î”ABC ≅ Î”PQR

**Answer**

Given,

AB = PQ, BC = QR and AM = PN

(i) 1/2 BC = BM and 1/2QR = QN (AM and PN are medians)

also,

BC = QR

⇒ 1/2 BC = 1/2QR

⇒ BM = QN

In Î”ABM and Î”PQN,

AM = PN (Given)

AB = PQ (Given)

BM = QN (Proved above)

Therefore, Î”ABM ≅ Î”PQN by SSS congruence condition.

(ii) In Î”ABC and Î”PQR,

AB = PQ (Given)

∠ABC = ∠PQR (by CPCT)

BC = QR (Given)

Therefore, Î”ABC ≅ Î”PQR by SAS congruence condition.

4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

**Answer**

BE and CF are two equal altitudes.

In Î”BEC and Î”CFB,

∠BEC = ∠CFB = 90° (Altitudes)

BC = CB (Common)

BE = CF (Common)

Therefore, Î”BEC ≅ Î”CFB by RHS congruence condition.

Now,

∠C = ∠B (by CPCT)

Thus, AB = AC as sides opposite to the equal angles are equal.

5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

**Answer**

AB = AC

In Î”ABP and Î”ACP,

∠APB = ∠APC = 90° (AP is altitude)

AB = AC (Given)

AP = AP (Common)

Therefore, Î”ABP ≅ Î”ACP by RHS congruence condition.

Thus, ∠B = ∠C (by CPCT)

Page No: 132

**Exercise 7.4**

1. Show that in a right angled triangle, the hypotenuse is the longest side.

**Answer**

Now,

∠A

⇒ ∠A + ∠C = 90° and ∠B is 90°.

Since, B is the largest angle of the triangle, the side opposite to it must be the largest.

So, BC is the hypotenuse which is the largest side of the right angled triangle ABC.

2. In Fig. 7.48, sides AB and AC of Î”ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC > AB.

**Answer**

Given,

∠PBC < ∠QCB

Now,

∠ABC + ∠PBC = 180°

⇒ ∠ABC = 180° - ∠PBC

also,

∠ACB + ∠QCB = 180°

⇒ ∠ACB = 180° - ∠QCB

Since,

∠PBC < ∠QCB therefore, ∠ABC > ∠ACB

Thus, AC > AB as sides opposite to the larger angle is larger.

3. In Fig. 7.49, ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

**Answer**

Given,

∠B < ∠A and ∠C < ∠D

Now,

AO < BO --- (i) (Side opposite to the smaller angle is smaller)

OD < OC ---(ii) (Side opposite to the smaller angle is smaller)

Adding (i) and (ii)

AO + OD < BO + OC

⇒ AD < BC

4. AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see Fig. 7.50).

Show that ∠A > ∠C and ∠B > ∠D.

**Answer**

In Î”ABD,

AB < AD < BD

∴ ∠ADB < ∠ABD --- (i) (Angle opposite to longer side is larger.)

Now,

In Î”BCD,

BC < DC < BD

∴ ∠BDC < ∠CBD --- (ii)

Adding (i) and (ii) we get,

∠ADB + ∠BDC < ∠ABD + ∠CBD

⇒ ∠ADC < ∠ABC

⇒ ∠B > ∠D

Similarly,

In Î”ABC,

∠ACB < ∠BAC --- (iii) (Angle opposite to longer side is larger.)

Now,

In Î”ADC,

∠DCA < ∠DAC --- (iv)

Adding (iii) and (iv) we get,

∠ACB + ∠DCA < ∠BAC + ∠DAC

⇒ ∠BCD < ∠BAD

⇒ ∠A > ∠C

5. In Fig 7.51, PR > PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

**Answer**

Given,

PR > PQ and PS bisects ∠QPR

To prove,

∠PSR > ∠PSQ

Proof,

∠PQR > ∠PRQ --- (i) (PR > PQ as angle opposite to larger side is larger.)

∠QPS = ∠RPS --- (ii) (PS bisects ∠QPR)

∠PSR = ∠PQR

∠PSQ = ∠PRQ + ∠RPS --- (iv) (exterior angle of a triangle equals to the sum of opposite interior angles)

Adding (i) and (ii)

∠PQR + ∠QPS > ∠PRQ + ∠RPS

⇒ ∠PSR > ∠PSQ [from (i), (ii), (iii) and (iv)]

Page No: 133

6. Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

**Answer**

Let

*l*is a line segment and B is a point lying o it. We drew a line AB perpendicular to

*l*. Let C be any other point on

*l*.

To prove,

AB < AC

Proof,

In Î”ABC,

∠B = 90°

Now,

∠A

⇒ ∠A + ∠C = 90°

∴ ∠C must be acute angle. or ∠C < ∠B

⇒ AB < AC (Side opposite to the larger angle is larger.)

Chapter 7 Triangles NCERT Solutions which will be helpful in revising the important theorems and topics. A closed figure formed by three intersecting lines is called a triangle. We will be studying the congruence of triangles, rules of congruence, some more properties of triangles and inequalities in a triangle.

• Congruence of Triangles: Two congruent figures have exactly the same shape and size. In congruent triangles corresponding parts are equal and we write in short ‘CPCT’ for corresponding parts of congruent triangles.

• Criteria for Congruence of Triangles:

SAS congruence rule- Two triangles are congruent if two sides and the included angle of one triangle are equal to the two sides and the included angle of the other triangle.

ASA congruence rule- Two triangles are congruent if two angles and the included side of one triangle are equal to two angles and the included side of other triangle. Two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. This is called the AAS Congruence Rule.

• Some Properties of a Triangle: Angles opposite to equal sides of an isosceles triangle are equaland the converse is the sides opposite to equal angles of a triangle are equal.

• Some More Criteria for Congruence of Triangles:

SSS congruence rule- If three sides of one triangle are equal to the three sides of another triangle, then the two triangles are congruent.

RHS congruence rule- If in two right triangles the hypotenuse and one side of one triangle are equal to the hypotenuse and one side of the other triangle, then the two triangles are congruent.

• Inequalities in a Triangle:

(i) If two sides of a triangle are unequal, the angle opposite to the longer side is larger (or greater).

(ii) In any triangle, the side opposite to the larger (greater) angle is longer.

(iii) The sum of any two sides of a triangle is greater than the third side.

There are total five exercises in which last one is optional. These

**Chapter 7 NCERT Solutions**are will increase your understanding of Triangles and will increase concentration among students. Below we have provided exercisewise NCERT Solutions which you can check.- Exercise 7.1 Chapter 7 Class 9 Maths NCERT Solutions
- Exercise 7.2 Chapter 7 Class 9 Maths NCERT Solutions
- Exercise 7.3 Chapter 7 Class 9 Maths NCERT Solutions
- Exercise 7.4 Chapter 7 Class 9 Maths NCERT Solutions
- Exercise 7.5 Chapter 7 Class 9 Maths NCERT Solutions

Our subject matter experts have prepared these NCERT Solutions through which one can clear their doubts and understand them easily.

### NCERT Solutions for Class 9 Maths Chapters:

**FAQ on**

**Chapter**

**7 Triangles**

#### How many exercises in Chapter 7 Triangles?

How many exercises in Chapter 7 Triangles?

Chapter 7 Triangles consists of total five exercises however one is optional not useful for the purpose of exams but will check your in depth knowledge. Here Studyrankers experts have provided accurate and detailed solutions of every question.

#### Each of the equal angles of an isosceles triangle is 38°, what is the measure of the third angle?

Each of the equal angles of an isosceles triangle is 38°, what is the measure of the third angle?

Let the third angle = x

∴ x + 38° + 38° = 180°

⇒ x = 180° - 38° - 38°

= 104Âº

#### Find the measure of each of acute angle in a right angle isosceles triangle.

Let the measure of each of the equal acute angle of the Î” be x

∴ We have: x + x + 90° = 180°

⇒ x + x = 180° - 90° = 90°

⇒ x= (90°/2)= 45°

#### If two angles are (30 ∠ a)Âº and (125 + 2a)Âº and they are supplement of each other. Find the value of ‘a’.

(30 - a)Âº and (125 + 2a)Âº are supplement to each other.

∴ (30 - a + 125 + 2a)Âº = 180Âº

⇒ a = 180Âº - 125Âº - 30Âº = 25Âº

⇒ Value of a = 25°