NCERT Solutions Class 9 Maths Chapter 7 Triangles : Exercise 7.2

Exercise 7.2 of NCERT Solutions for Class 9 Maths Chapter 7 covers the properties of triangles, particularly the isosceles triangle.  This exercise introduces and uses important theorems regarding isosceles triangles: first, the angles opposite to equal sides of an isosceles triangle are equal; and second, it conversely shows that sides opposite equal angles of a triangle are equal. These properties are critical for solving problems in which certain sides or angles have been described with respect to an isosceles triangle.

Exercise 7.2 also often correlates with additional congruence criteria in several other congruence situations particularly ASA and AAS (Angle-Angle-Side) in different circumstances to show or prove a variety of relationships in triangles. This allows students to further their understanding of congruence through different problem contexts.

1.0Download NCERT Solutions of Class 9 Maths Chapter 7 Triangles Exercise 7.2 : Free PDF

Download the NCERT Solutions Class 9 Maths Chapter 7, Triangles Exercise 7.2  PDF file to study offline. You can practice with these questions over and over again!

Key Concepts of Exercise 7.2

• Concept of Triangle Congruence
• Triangle Congruence Postulates: SSS, SAS, ASA, AAS
• Using the congruence criteria to show that two triangles are congruent
• Using congruence to deduce equal parts of triangles
• Providing logical reasoning based on the idea that we are given information and using the diagram

2.0NCERT Exercise Solutions Class 9 Chapter 7 Triangles: All Exercises 

Here is a quick overview and access to solutions for all exercises from Chapter 7, Triangles.

3.0NCERT Class 9 Maths Chapter 7 Exercise 7.2 : Detailed Solutions

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.

Sol.

Given: ΔABC is an isosceles triangle with AB=AC. OB bisects ∠B, OC bisects ∠C. OB and OC intersect at O.

To prove:

(i) OB=OC

(ii) AO bisects ∠A.

Proof:

(i) In ΔABC, since AB=AC (Given), the angles opposite to these sides are equal.

Therefore, ∠ABC = ∠ACB.

Since OB bisects ∠B, ∠OBC = (1/2)∠ABC.

Since OC bisects ∠C, ∠OCB = (1/2)∠ACB.

Since ∠ABC = ∠ACB, it follows that (1/2)∠ABC = (1/2)∠ACB.

Therefore, ∠OBC = ∠OCB.

Now, in ΔOBC, since ∠OBC = ∠OCB, the sides opposite to these equal angles are equal.

Thus, OB=OC.

(ii) In ΔOAB and ΔOAC:

1. AB = AC (Given)
2. OB = OC (Proved in (i))
3. AO = AO (Common side)
   Therefore, by SSS (Side-Side-Side) congruence criteria, ΔOAB≅ΔOAC.
   By CPCT, ∠OAB = ∠OAC.
   Thus, AO bisects ∠A.

2. In ΔABC, AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB=AC.

Sol.

Given: In ΔABC, AD is the perpendicular bisector of BC. This means AD⊥BC and DB=DC.

To prove: ΔABC is an isosceles triangle with AB=AC.

Proof: In ΔADB and ΔADC:

1. DB = DC (Given, AD is the bisector of BC)
2. ∠ADB = ∠ADC (Each = 90°, given AD is perpendicular to BC)
3. AD = AD (Common side)
   Therefore, by SAS (Side-Angle-Side) congruence rule, ΔADB≅ΔADC.
   By CPCT, AB=AC.
   Since two sides of ΔABC are equal (AB=AC), ΔABC is an isosceles triangle.

3. ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.

Sol.

Given: ΔABC is an isosceles triangle with AB=AC. BE⊥AC and CF⊥AB.

To prove: BE=CF.

Proof: In ΔABE and ΔACF:

1. ∠A = ∠A (Common angle)
2. ∠AEB = ∠AFC (Each = 90°, since BE⊥AC and CF⊥AB)
3. AB = AC (Given, ΔABC is an isosceles triangle)
   Therefore, by AAS (Angle-Angle-Side) congruence criteria, ΔABE≅ΔACF.
   By CPCT, BE=CF.
   Hence, the altitudes are equal.

4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (See figure). Show that

(i) ΔABE≅ΔACF

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

Sol.

Given: In ΔABC, BE⊥AC and CF⊥AB. Also, BE=CF.

To prove:

(i) ΔABE≅ΔACF

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

Proof:

(i) In ΔABE and ΔACF:

1. ∠A = ∠A (Common angle)
2. ∠AEB = ∠AFC (Each = 90°, since BE⊥AC and CF⊥AB)
3. BE = CF (Given)
   Therefore, by AAS (Angle-Angle-Side) congruence criteria, ΔABE≅ΔACF.

(ii) Since ΔABE≅ΔACF (Proved above), by CPCT, AB=AC.

Since two sides of ΔABC are equal (AB=AC), ΔABC is an isosceles triangle.

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

Sol.

Given: ΔABC is an isosceles triangle with AB=AC on base BC. ΔDBC is an isosceles triangle with DB=DC on base BC.

To prove: ∠ABD=∠ACD.

Proof:

In ΔABC, since AB=AC, the angles opposite to these sides are equal.

Therefore, ∠ABC = ∠ACB. ... (1)

In ΔDBC, since DB=DC, the angles opposite to these sides are equal.

Therefore, ∠DBC = ∠DCB. ... (2)

Add equation (1) and equation (2):

∠ABC + ∠DBC = ∠ACB + ∠DCB

This implies ∠ABD = ∠ACD.

6. ΔABC is an isosceles triangle in which AB=AC. Side BA is produced to D such that AD=AB (see figure). Show that ∠BCD is a right angle.

Sol.

Given: ΔABC is an isosceles triangle with AB=AC. BA is produced to D such that AD=AB.

To prove: ∠BCD is a right angle (∠BCD=90°).

Proof:

In ΔABC, since AB=AC (Given), the angles opposite to these sides are equal.

∠ACB = ∠ABC. ... (1)

We are given AD=AB. Since AB=AC, by transitivity, AD=AC.

In ΔACD, since AD=AC, the angles opposite to these sides are equal.

∠ACD = ∠ADC. ... (2)

Now, consider ΔBCD. The sum of its angles is 180°.

∠DBC + ∠BCD + ∠CDB = 180°

Replace ∠DBC with ∠ABC and ∠CDB with ∠ADC.

∠ABC + ∠BCD + ∠ADC = 180°

From (1), ∠ABC = ∠ACB.

From (2), ∠ADC = ∠ACD.

Substitute these into the angle sum:

∠ACB + ∠BCD + ∠ACD = 180°

Since ∠BCD = ∠ACB + ∠ACD, we can write:

∠BCD + ∠BCD = 180°

2∠BCD = 180°

∠BCD = 180°/2

Therefore, ∠BCD = 90°.

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

Sol.

Given: ΔABC is a right-angled triangle with ∠A = 90° and AB=AC.

To find: ∠B and ∠C.

Solution:

In ΔABC, since AB=AC (Given), the angles opposite to these equal sides are equal.

Therefore, ∠B = ∠C. ... (1)

The sum of angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°

Given ∠A = 90°.

90° + ∠B + ∠C = 180°

∠B + ∠C = 180° − 90°

∠B + ∠C = 90°. ... (2)

From (1) and (2), substitute ∠C with ∠B:

∠B + ∠B = 90°

2∠B = 90°

∠B = 90°/2 = 45°.

Since ∠B = ∠C, then ∠C = 45°.

Therefore, ∠B = 45° and ∠C = 45°.

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

Sol.

Given: ΔABC is an equilateral triangle.

To prove: ∠A = ∠B = ∠C = 60°.

Proof:

Since ΔABC is an equilateral triangle, all its sides are equal.

AB = BC = CA.

Since AB = BC, the angles opposite to these sides are equal.

Therefore, ∠C = ∠A. ... (1)

Similarly, since BC = CA, the angles opposite to these sides are equal.

Therefore, ∠A = ∠B. ... (2)

From (1) and (2), we have ∠A = ∠B = ∠C. ... (3)

The sum of angles in a triangle is 180°.

∠A + ∠B + ∠C = 180°. ... (4)

Substitute ∠A for ∠B and ∠C from (3) into (4):

∠A + ∠A + ∠A = 180°

3∠A = 180°

∠A = 180°/3 = 60°.

Since ∠A = ∠B = ∠C,

Therefore, ∠A = ∠B = ∠C = 60° each.

4.0Key Features and Benefits Class 9 Maths Chapter 7 Triangles: Exercise 7.2

• Comprehensive Concept Coverage: Each Solution describes which rule is being used and why.
• Logical Approach: Encourages students to develop logical proof-writing processes through this method of learning.
• Visual Approaches: Diagrams and visual prompts will help students understand relationships.
• CBSE Mentor Maps: Each exercise is derived from the NCERT textbook and the patterns you will encounter in exams.
• Great for Practice & Revision: Excellent starting points for reinforcing congruence concepts before examination