NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2

Vishal kumarUpdated on 21 May 2025, 03:34 PM IST

The exercise uses fundamental geometry principles to explore important characteristics of isosceles triangles. The tasks demand to validate statements regarding angle bisectors as well as altitudes and side lengths by using congruence rules including SAS, ASA and RHS. The problems enables to develop a stronger knowledge of triangle congruence alongside its features. All diagrams in the textbook along with their explanatory text assist us to understand how congruent parts function, reinforcing important geometric patterns and reasoning.

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NCERT Solutions for Class 9 Maths Chapter 7 – Triangles Exercise 7.2
Access Solution of Triangles Class 9 Chapter 7 Exercise: 7.2
Topics Covered in Chapter 7 Triangles: Exercise 7.2
NCERT Solutions of Class 10 Subject Wise
Subject Wise NCERT Exemplar Solutions

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

Students must establish equality between line segments or angles when they apply necessary conditions using logical reasoning and axioms. The practice problems serve multiple goals because they develop proficiency in proof writing while students learn advanced geometry connections. Regular use of NCERT Solutions for reinforcement purposes supports students in learning triangle concepts while NCERT Books provide complete understanding through practice.

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

Access Solution of Triangles Class 9 Chapter 7 Exercise: 7.2

Q1 (i) 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 : OB=OC

Answer:

Given, AB = AC, and ABC is an isosceles triangle. 1000005876

Therefore, it means ∠B=∠C

or 12∠B = 12∠C

Thus, ∠OBC = ∠OCB (Angles bisectors are equal)

Hence, OB=OC as sides opposite to equal are angles are also equal.

Q1 (ii) 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 : AO bisects ∠A

Answer:

1000005876

Consider ΔAOB and ΔAOC ,

(i) AB = AC (Given)

(ii) AO = AO (Common in both the triangles)

(iii) OB = OC (Proved in previous part)

Therefore by SSS congruence rule, we can conclude that :

ΔAOB ≅ ΔAOC

Now, by c.p.c.t.,

∠BAO = ∠CAO

Hence AO bisects ∠A .

Q2 In ΔABC , AD is the perpendicular bisector of BC (see Fig). Show that ΔABC is an isosceles triangle in which AB=AC .

1640157417015

Answer:

Given: AD is perpendicular bisector of BC.

To Prove: AB = AC

Proof: Consider Δ ABD and Δ ADC,

(i) AD = AD (Common in both the triangles)

(ii) ∠ADB = ∠ADC (Right angle)

(iii) BD = CD ( as AD is perpendicular to side BC)

Therefore, by SAS congruence criteria:

ΔADB ≅ ΔADC

Thus, AB = AC ( by c.p.c.t )

Hence Proved

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

1640165152512

Answer:

Given: AC = AB and BE and CF are altitudes.

To Prove: BE = CF

Proof: Consider ΔAEB and ΔAFC ,

(i) ∠A is common in both the triangles.

(ii) ∠AEB = ∠AFC (Right angles)

(iii) AB = AC (Given)

Thus by AAS congruence axiom, we can conclude that :

ΔAEB ≅ΔAFC

Now, by c.p.c.t. we can say : BE = CF

Hence Proved

Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that ΔABE≅ΔACF

1640165171491

Answer:

Given: BE = CF

Consider ΔABE and ΔACF ,

(i) ∠A is common in both the triangles.

(ii) ∠AEB = ∠AFC (Right angles)

(iii) BE = CF (Given)

Thus by AAS congruence, we can say that :

ΔABE≅ΔACF

Hence Proved

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that AB=AC , i.e., ABC is an isosceles triangle.

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Answer:

From the prevoius part of the question we found out that : ΔABE ≅ΔACF

Now, by c.p.c.t. we can say that : AB = AC

Thus, Δ ABC is an isosceles triangle.

Hence Proved

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that ∠ABD =∠ACD.

1640166831546

Answer:

Given: ABC and DBC are isosceles triangles and

To Prove: ∠ABD =∠ACD

Proof: Consider ΔABD and ΔACD,

(i) AD = AD (Common in both the triangles)

(ii) AB = AC (Sides of isosceles triangle)

(iii) BD = CD (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that:

△ABD ≅ triangleACD

Therefore, by c.p.c.t.,

∠ABD =∠ACD

Hence Proved

Q6 ΔABC is an isosceles triangle in which AB=AC . Side BA is produced to D such that AD=AB (see Fig.). Show that ∠BCD is a right angle.

1640157465047

Answer:

Given: AB = AC and AD = AB

To Prove: ∠BCD is a right angle

Proof: Consider Δ ABC,
It is given that AB = AC
So, ∠ACB=∠ABC (Since angles opposite to the equal sides are equal.)

Similarly in Δ ACD, AD = AB and ∠ADC=∠ACD
So,

∠CAB+∠ACB+∠ABC=180∘

∠CAB + 2∠ACB=180∘
or ∠CAB =180∘ − 2∠ACB ...........................(i)

And in Δ ADC,
∠CAD =180∘ − 2∠ACD ..............................(ii)

Adding (i) and (ii), we get :
∠CAB + ∠CAD =360∘ − 2∠ACD − 2∠ACB

or 180∘ =360∘ − 2∠ACD − 2∠ACB

and ∠BCD = 90∘

Hence Proved

Q7 ABC is a right angled triangle in which ∠A=90∘ and AB=AC . Find ∠B and ∠C .

Answer:

Given: AB = AC and ∠A=90∘

We know that angles opposite to equal sides are also equal.

Therefore, ∠B = ∠C

Also, the sum of the interior angles of a triangle is 180∘ .

So, we have :

∠A + ∠B + ∠C = 180∘

or 90∘+ 2∠B = 180∘

or ∠B = 45∘

As, ∠B = ∠C

So, angleC =\ 45^{\circ}$

Q8 Show that the angles of an equilateral triangle are 60∘ each.

Answer:

Consider a triangle ABC which has all sides equal as shown in the figure.

1000005878

We know that angles opposite to equal sides are equal.

Therefore: ∠A = ∠B = ∠C

Also, the sum of the interior angles of a triangle is 180∘ .

Hence, ∠A + ∠B + ∠C = 180∘

or 3∠A = 180∘

or ∠A = 60∘

As, all the angles of the equilateral triangle are equal, thus ∠A = ∠B = ∠C = 60∘.

Also Read:

Topics Covered in Chapter 7 Triangles: Exercise 7.2

Congruence of triangles: Congruence refers to the exact matching of size and shape between two triangles. It allows us to prove that corresponding sides and angles are equal using specific rules.
Properties of isosceles triangles: They exhibit two identical sides along with two congruent base angles. Geometric solutions and congruence proofs depend heavily on these characteristics of the isosceles triangle.
Use of perpendicular bisectors, angle bisectors, and altitudes: Triangles can be divided into equal or right-angled pieces with the use of these line segments. They are helpful in establishing congruent pieces and building evidence.
Application of RHS, SSS, and ASA congruence criteria: These rules—Side-Side-Side (SSS), Angle-Side-Angle (ASA), and Right Angle-Hypotenuse-Side (RHS)—are used to demonstrate that two triangles are congruent given the measurements.

Also, See

NCERT Solutions of Class 10 Subject Wise

Subject Wise NCERT Exemplar Solutions

Frequently Asked Questions (FAQs)

Q: Can we consider an equilateral triangle as an isosceles triangle?

Yes, every equilateral triangle is an isosceles triangle, but every isosceles triangle is not an equilateral triangle.

Q: The ratio in which the angle bisector of the vertex angle divides the base is?

The angle bisector of the vertex angle divides the base into 1:1 ratio which means into two equal parts i.e., it passes through the midpoint of the base.

Q: What is the definition of an isosceles triangle?

An isosceles triangle is one with two equal sides. It has two equal angles as well.

Q: How many sides and angles of an isosceles triangle are equal?

An isosceles triangle has two sides and two angles equal to each other.

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