CBSE Class 9 Maths Chapter 7 Triangles NCERT Solutions

Reviewed by NCERT Help Editorial Team · Updated on 29 Jul 2026

NCERT Solutions PDF Class 9 PDF

CBSE Class 9 Mathematics Chapter 7 introduces the essential topic of triangles, with a special emphasis on congruence. This chapter explores the conditions that determine if two triangles are identical in shape and size, commonly known as congruence criteria. These include the Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), and Right-Hand-Side (RHS) postulates. The NCERT Solutions for this chapter offer detailed, step-by-step guidance for all exercises. They illustrate how to effectively apply these congruence rules to prove various properties of triangles and tackle a wide range of geometric problems. Students will gain proficiency in proving triangle congruence and utilizing the concept of Corresponding Parts of Congruent Triangles (CPCT) to demonstrate the equality of sides and angles. This foundational knowledge is crucial for advanced geometry and exam preparation.

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Quick info

Board	CBSE
Class	Class 9
Subject	Mathematics
Session	2026
Language	English
Type	NCERT Solutions
Chapter	Chapter 7

Chapter summary

Chapter 7, Triangles, focuses on the core concept of triangle congruence. The NCERT Solutions provide detailed explanations for proving congruence using criteria like SAS, AAS, and ASA. Students will practice applying these rules to solve problems involving quadrilaterals, angle bisectors, and perpendiculars. The solutions emphasize the use of CPCT to deduce equality of sides and angles, reinforcing understanding of geometric proofs. This chapter is crucial for developing logical reasoning and problem-solving skills in geometry.

Learning outcomes

Understand the conditions for triangle congruence (SSS, SAS, ASA, AAS).
Apply congruence criteria to prove that two triangles are congruent.
Utilize CPCT (Corresponding Parts of Congruent Triangles) to find equal sides and angles.
Solve problems involving geometric figures using triangle congruence.
Prove that a line segment bisects another line segment.
Demonstrate understanding of angle bisectors and perpendiculars in geometric proofs.

Topics covered

Paper topics

Introduction to Triangles
Congruence of Triangles
Criteria for Congruence (SSS, SAS, ASA, AAS)
Corresponding Parts of Congruent Triangles (CPCT)
Proving Equality of Sides and Angles
Geometric Proofs
Angle Bisectors
Perpendiculars
Properties of Quadrilaterals
Line Segments and Parallel Lines

Important topics

Criteria for Congruence (SSS, SAS, ASA, AAS)
Corresponding Parts of Congruent Triangles (CPCT)
Applying Congruence to Prove Geometric Statements
Solving Problems involving Angle Bisectors and Perpendiculars
Proving Equality of Sides and Angles using Congruence

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Questions and Solutions

Question 1

In quadrilateral ACBD, $AC = AD$ and AB bisects $\angle A$ (see Fig.). Show that $\triangle ABC \cong \triangle ABD$ . What can you say about BC and BD?

Solution:

To show that $\triangle ABC \cong \triangle ABD$ , we need to find three conditions of congruence. Let's analyze the given information:

$AC = AD$ (Given)
$\angle CAB = \angle DAB$ (Since AB bisects $\angle A$ , it divides the angle into two equal parts).
$AB = AB$ (This side is common to both triangles).

By the Side-Angle-Side (SAS) congruence criterion, since two sides and the included angle of $\triangle ABC$ are equal to the corresponding two sides and the included angle of $\triangle ABD$ , we can conclude that:

$\triangle ABC \cong \triangle ABD$

Now, we need to determine the relationship between BC and BD. Since the triangles are congruent, their corresponding parts are equal (CPCT - Corresponding Parts of Congruent Triangles).

Therefore, $BC = BD$ .

Answer: $\triangle ABC \cong \triangle ABD$ by SAS congruence, and $BC = BD$ .

Question 2

ABCD is a quadrilateral in which $AD = BC$ and $\angle DAB = \angle CBA$ (see Fig.). Prove that (i) $\triangle ABD \cong \triangle BAC$ (ii) $BD = AC$ (iii) $\angle ABD = \angle BAC$

Solution:

We are given a quadrilateral ABCD with $AD = BC$ and $\angle DAB = \angle CBA$ . We need to prove three statements.

Part (i): Prove $\triangle ABD \cong \triangle BAC$

Consider $\triangle ABD$ and $\triangle BAC$ . We have:

$AD = BC$ (Given)
$\angle DAB = \angle CBA$ (Given)
$AB = AB$ (Common side)

By the Side-Angle-Side (SAS) congruence criterion, since two sides and the included angle of $\triangle ABD$ are equal to the corresponding two sides and the included angle of $\triangle BAC$ , we have:

$\triangle ABD \cong \triangle BAC$

Part (ii): Prove $BD = AC$

Since $\triangle ABD \cong \triangle BAC$ (proved in part (i)), their corresponding parts are equal (CPCT).

Therefore, $BD = AC$ .

Part (iii): Prove $\angle ABD = \angle BAC$

Again, since $\triangle ABD \cong \triangle BAC$ (proved in part (i)), their corresponding parts are equal (CPCT).

Therefore, $\angle ABD = \angle BAC$ .

Proved.

Question 3

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

Solution:

We are given that AD and BC are equal perpendiculars to the line segment AB. This means $AD \perp AB$ , $BC \perp AB$ , and $AD = BC$ . We need to show that CD bisects AB, which means we need to prove that the point of intersection of CD and AB is the midpoint of AB.

Let O be the point of intersection of CD and AB. Consider $\triangle AOD$ and $\triangle BOC$ .

We have the following information:

$\angle DAO = \angle CBO$ (Each is equal to $90^{\circ}$ because AD and BC are perpendiculars to AB).
$AD = BC$ (Given)
$\angle AOD = \angle BOC$ (Vertically opposite angles are equal).

By the Angle-Angle-Side (AAS) congruence criterion, since two angles and a non-included side of $\triangle AOD$ are equal to the corresponding two angles and the non-included side of $\triangle BOC$ , we have:

$\triangle AOD \cong \triangle BOC$

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Therefore, $AO = BO$ .

This means that the point O is the midpoint of the line segment AB. Hence, CD bisects AB.

Proved.

Question 4

l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig.). Show that $\triangle ABC \cong \triangle CDA$ .

Solution:

We are given two pairs of parallel lines, l || m and p || q. Let the lines p and q intersect l and m at points A, B, C, and D respectively, forming a quadrilateral ABCD. AC is a diagonal. We need to show that $\triangle ABC \cong \triangle CDA$ .

Since p || q, we can consider line AC as a transversal. Therefore, the alternate interior angles are equal:

$\angle BAC = \angle DCA$

Since l || m, we can again consider line AC as a transversal. Therefore, the alternate interior angles are equal:

$\angle BCA = \angle DAC$

Now, consider the two triangles $\triangle ABC$ and $\triangle CDA$ .

We have:

$\angle BAC = \angle DCA$ (Alternate angles)
$AC = AC$ (Common side)
$\angle BCA = \angle DAC$ (Alternate angles)

By the Angle-Side-Angle (ASA) congruence criterion, since two angles and the included side of $\triangle ABC$ are equal to the corresponding two angles and the included side of $\triangle CDA$ , we have:

$\triangle ABC \cong \triangle CDA$

Proved.

Question 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 $\angle A$ (see Fig.). Show that : (i) $\triangle APB \cong \triangle AQB$ (ii) $BP = BQ$ or B is equidistant from the arms of $\angle A$ .

Solution:

We are given that line l bisects $\angle A$ , and B is a point on l. BP and BQ are perpendiculars from B to the arms of $\angle A$ . This means $\angle APB = 90^{\circ}$ and $\angle AQB = 90^{\circ}$ .

Part (i): Show that $\triangle APB \cong \triangle AQB$

Consider $\triangle APB$ and $\triangle AQB$ . We have:

$\angle PAB = \angle QAB$ (Since l is the bisector of $\angle A$ ).
$AB = AB$ (Common side)
$\angle APB = \angle AQB$ (Each is equal to $90^{\circ}$ as BP and BQ are perpendiculars).

By the Angle-Angle-Side (AAS) congruence criterion, since two angles and a non-included side of $\triangle APB$ are equal to the corresponding two angles and the non-included side of $\triangle AQB$ , we have:

$\triangle APB \cong \triangle AQB$

Part (ii): Show that $BP = BQ$

Since $\triangle APB \cong \triangle AQB$ (proved in part (i)), their corresponding parts are equal (CPCT).

Therefore, $BP = BQ$ .

This means that point B is equidistant from the arms of $\angle A$ .

Proved.

Question 6

In the figure, $AC = AE$ , $AB = AD$ and $\angle BAD = \angle EAC$ . Show that $BC = DE$ .

Solution:

We are given that $AC = AE$ , $AB = AD$ , and $\angle BAD = \angle EAC$ . We need to show that $BC = DE$ .

First, let's establish a relationship between the angles $\angle BAC$ and $\angle DAE$ .

We are given: $\angle BAD = \angle EAC$

Add $\angle DAC$ to both sides of the equation:

$\angle BAD + \angle DAC = \angle EAC + \angle DAC$

Observing the figure, we can see that:

$\angle BAD + \angle DAC = \angle BAC$

$\angle EAC + \angle DAC = \angle DAE$

Therefore, we can conclude that:

$\angle BAC = \angle DAE$ ... (i)

Now, let's consider the triangles $\triangle ABC$ and $\triangle ADE$ .

We have the following information:

$AB = AD$ (Given)
$AC = AE$ (Given)
$\angle BAC = \angle DAE$ (From equation (i))

$\triangle ABC \cong \triangle ADE$

Since the triangles are congruent, their corresponding parts are equal (CPCT).

Therefore, $BC = DE$ .

Proved.

Common mistakes

Incorrectly identifying corresponding vertices, sides, or angles when applying congruence rules.
Confusing the order of vertices in congruence statements (e.g., writing \triangle ABC \cong \triangle BAC when it should be \triangle ABC \cong \triangle ABD).
Errors in applying the SAS, ASA, or AAS congruence criteria due to misinterpreting given information.
Forgetting to state the reason for each step in a geometric proof (e.g., Given, Common, CPCT, Alternate angles).
Misapplying the concept of vertically opposite angles or alternate interior angles.

Revision tips

Clearly identify the two triangles you need to prove congruent for each problem.
List the given information and common sides/angles first.
Carefully match the corresponding sides and angles based on the given information and the congruence rule being used.
Always state the reason (e.g., Given, Common, CPCT, SAS, ASA) for each step in your proof.
Practice drawing the figures accurately to visualize the relationships between different parts of the triangles.

Practice MCQs

Q1. In $\triangle$ ABC and $\triangle$ ABD, if A $C = AD and AB bisects$ $\angle$ A, which congruence rule can be used to prove $\triangle$ ABC $\cong$ $\triangle$ ABD?

Q2. If two triangles are congruent, what can be said about their corresponding sides and angles?

Q3. In quadrilateral ABCD, if A $D = BC and$ $\angle$ DA $B = \angle$ CBA, which congruence rule proves $\triangle$ ABD $\cong$ $\triangle$ BAC?

Q4. What does it mean for CD to bisect AB?

Q5. Which congruence rule is used to prove $\triangle$ AOD $\cong$ $\triangle$ BOC in Q.3, given A $D = BC and$ $\angle$ DA $O = \angle$ CB $O = 90^{$ $\circ$ }?

Frequently asked questions

What is the main focus of Chapter 7: Triangles in Class 9 Maths?

Chapter 7 primarily focuses on the concept of congruence of triangles. It explains the different criteria (SSS, SAS, ASA, AAS) used to prove that two triangles are congruent and how to use this congruence to prove other properties.

What does CPCT stand for and why is it important?

CPCT stands for Corresponding Parts of Congruent Triangles. It is a crucial principle that states if two triangles are congruent, then all their corresponding sides and corresponding angles are equal. It's used to derive results after proving congruence.

How do these NCERT Solutions help students?

These solutions provide clear, step-by-step explanations for each problem in Chapter 7. They help students understand the logic behind geometric proofs, practice applying congruence rules, and build confidence for exams.

What are the key congruence rules covered in this chapter?

The key congruence rules covered are Side-Side-Side (SSS), Side-Angle-Side (SAS), Angle-Side-Angle (ASA), and Angle-Angle-Side (AAS).

Are the figures provided in the NCERT textbook used in the solutions?

Yes, the solutions refer to the figures provided in the NCERT textbook to help visualize the problem and understand the geometric relationships involved.

How can I use these solutions to prepare for my exams?

You can use these solutions to understand the methods for solving problems, practice the steps involved in proofs, and check your own answers. Reworking the problems after understanding the solution is highly recommended.

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CBSE Class 9 Maths Chapter 7 Triangles NCERT Solutions

Quick info

Chapter summary

Learning outcomes

Topics covered

Paper topics

Important topics

PDF preview

Questions and Solutions

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Common mistakes

Revision tips

Practice MCQs

Frequently asked questions

Important links