 # NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2

Exercise 7.2 of Class 9 Maths Chapter 7 consists of properties of Triangles and Theorems related to it. Most students regard this topic as boring as it is more of proving and less of solving. But, if students understand the properties of Triangles by correlating it to the examples then they will start taking an interest in it. Here, we have provided the NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.2 for students convenience. After going through the solutions, students can easily understand the method of solving the questions.

The NCERT Solutions for Class 9 Maths Chapter 7 Triangles is solved by our team of experienced teachers. These solutions will help students to understand the way of approaching the questions. Also, it will help them write the answers in a step by step format, so that they can score high marks in the annual Maths paper. To download the Exercise 7.2 solution PDF, visit the link below.

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The chapter 7 consists of a total of 5 exercises. To get the solutions of other exercise visit the link below.

Exercise 7.1 Solution 8 Questions (6 Short Answer Questions, 2 Long Answer Question)

Exercise 7.3 Solution 5 Questions (3 Short Answer Questions, 2 Long Answer Question)

Exercise 7.4 Solution 6 Questions (5 Short Answer Questions, 1 Long Answer Question)

Exercise 7.5 (Optional) Solution 4 Questions

### Access Answers to Chapter 7 – Triangles 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 Solution:

Given:

AB = AC and

the bisectors of B and C intersect each other at O

(i) Since ABC is an isosceles with AB = AC,

B = C

½ B = ½ C

⇒ OBC = OCB (Angle bisectors)

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

(ii) In ΔAOB and ΔAOC,

AB = AC (Given in the question)

AO = AO (Common arm)

OB = OC (As Proved Already)

So, ΔAOB ΔAOC by SSS congruence condition.

BAO = CAO (by CPCT)

Thus, AO bisects A.

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

It is given that AD is the perpendicular bisector of BC

To prove:

AB = AC

Proof:

BD = CD (Since AD is the perpendicular bisector)

Thus,

AB = AC (by CPCT)

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:

(i) BE and CF are altitudes.

(ii) AC = AB

To prove:

BE = CF

Proof:

Triangles ΔAEB and ΔAFC are similar by AAS congruency since

A = A (It is the common arm)

AEB = AFC (They are right angles)

AB = AC (Given in the question)

∴ ΔAEB ΔAFC and so, 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) ΔABE ΔACF

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

It is given that BE = CF

(i) In ΔABE and ΔACF,

A = A (It is the common angle)

AEB = AFC (They are right angles)

BE = CF (Given in the question)

∴ ΔABE ΔACF by AAS congruency condition.

(ii) AB = AC by CPCT and so, 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. Solution:

In the question, it is given that ABC and DBC are two isosceles triangles.

We will have to show that ABD = ACD

Proof:

Triangles ΔABD and ΔACD are similar by SSS congruency since

AB = AC (Since ABC is an isosceles triangle)

BD = CD (Since BCD is an isosceles triangle)

So, ΔABD ΔACD.

∴ ABD = ACD by CPCT.

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:

It is given that AB = AC and AD = AB

We will have to now prove BCD is a right angle.

Proof:

Consider ΔABC,

AB = AC (It is given in the question)

Also, ACB = ABC (They are angles opposite to the equal sides and so, they are equal)

Now, consider ΔACD,

Also, ADC = ACD (They are angles opposite to the equal sides and so, they are equal)

Now,

In ΔABC,

CAB + ACB + ABC = 180°

So, CAB + 2ACB = 180°

⇒ CAB = 180° – 2ACB — (i)

CAD = 180° – 2ACD — (ii)

also,

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

Adding (i) and (ii) we get,

CAB + CAD = 180° – 2ACB+180° – 2ACD

⇒ 180° = 360° – 2ACB-2ACD

⇒ 2(ACB+ACD) = 180°

⇒ BCD = 90°

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

Solution: In the question, it is given that

A = 90° and AB = AC

AB = AC

⇒ B = C (They are angles opposite to the equal sides and so, they are equal)

Now,

A+B+C = 180° (Since the sum of the interior angles of the triangle)

∴ 90° + 2B = 180°

⇒ 2B = 90°

⇒ B = 45°

So, B = C = 45°

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

Solution:

Let ABC be an equilateral triangle as shown below: Here, BC = AC = AB (Since the length of all sides is same)

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

Also, we know that

A+B+C = 180°

⇒ 3A = 180°

⇒ A = 60°

∴ A = B = C = 60°

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

This exercise covers the properties of Triangles along with the two theorems as mentioned below:

1. Theorem 7.2 : Angles opposite to equal sides of an isosceles triangle are equal.
2. Theorem 7.3: The sides opposite to equal angles of a triangle are equal.

These theorems can be proved by the congruence rule that students have studied in previous exercise i.e 7.1. So, go through the proofs in depth to understand it. Also, the questions provided in the exercise are based on these theorems. Students should try to solve the questions by themselves. If they get stuck somewhere then they can refer to the solutions. These solutions are provided to help students in their studies so that they get rid of all their doubts.

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