• NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • Class 6-10
      • Class 6th
      • Class 7th
      • Class 8th
      • Class 9th
      • Class 10th
    • View All Options
      • Online Courses
      • Distance Learning
      • Hindi Medium Courses
      • International Olympiad
    • NEET
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE (Main+Advanced)
      • Class 11th
      • Class 12th
      • Class 12th Plus
    • JEE Main
      • Class 11th
      • Class 12th
      • Class 12th Plus
  • Classroom
  • NEW
    • NEET
      • 2025
      • 2024
      • 2023
      • 2022
    • JEE
      • 2025
      • 2024
      • 2023
      • 2022
    • Class 6-10
    • JEE Main
      • Previous Year Papers
      • Sample Papers
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • JEE Advanced
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
    • NEET
      • Previous Year Papers
      • Sample Papers
      • Mock Test
      • Result
      • Analysis
      • Syllabus
      • Exam Date
      • College Predictor
      • Counselling
    • NCERT Solutions
      • Class 6
      • Class 7
      • Class 8
      • Class 9
      • Class 10
      • Class 11
      • Class 12
    • CBSE
      • Notes
      • Sample Papers
      • Question Papers
    • Olympiad
      • NSO
      • IMO
      • NMTC
    • TALLENTEX
    • AOSAT
  • ALLEN E-Store
    • ALLEN for Schools
    • About ALLEN
    • Blogs
    • News
    • Careers
    • Request a call back
    • Book home demo
NCERT Solutions
Class 9
Maths
Chapter 7 Triangles
Exercise 7.1

NCERT Solutions Class 9 Maths Chapter 7 Triangles : Exercise 7.1

Triangles are fundamental figures in geometry, forming the basis for many complex mathematical concepts. In NCERT Solutions Class 9 Maths Chapter 7 Triangles, students dive deep into triangle properties and congruence criteria that help solve problems with precision and logical reasoning.

NCERT Solutions Class 9 Maths Chapter 7 Triangles Exercise 7.1 specifically deals with understanding the criteria for the congruence of triangles.Mastering these concepts builds a strong foundation for geometry in higher classes and competitive exams.

1.0Download NCERT Solutions of Class 9 Maths Chapter 7 Triangles Exercise 7.1 : Free PDF

Get reliable and simplified NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 in PDF format. Downloading these solutions will help students to understand and work through the various congruence rules, step by step, and develop the skills necessary for problem solving.

NCERT Solutions Class 9 Maths Chapter 7 Ex 7.1

Key Concepts of Exercise 7.1

  • Understanding congruence of triangles and methods of establishing them . 
  • Implementing the SSS, SAS, ASA, RHS methods of proving triangles congruent. 
  • Solving congruence questioning style (proof based), as well as numerical. 
  • Applying congruence strategies using real-life geometric content. 
  • Building inference and deduction skills using geometrical content.

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.

NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.1

NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.2

NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.3

3.0NCERT Class 9 Maths Chapter 7 Exercise 7.1 : Detailed Solutions

1. In quadrilateral ACBD, AC=AD and AB bisects ∠A. Show that ΔABC≅ΔABD. What can you say about BC and BD?

Geometry of Triangle Image 8

Sol.

Given: In quadrilateral ACBD, AC=AD and AB bisects ∠A.

To prove: ΔABC≅ΔABD.

Proof: In ΔABC and ΔABD:

  1. AC = AD (Given)
  2. AB = AB (Common side)
  3. ∠CAB = ∠DAB (Since AB bisects ∠A)
    Therefore, by SAS (Side-Angle-Side) congruence criteria, ΔABC≅ΔABD.

By CPCT (Corresponding Parts of Congruent Triangles), we can say that BC = BD.

2. ABCD is a quadrilateral in which AD=BC and ∠DAB=∠CBA. Prove that:

Geometry of Triangle Image 7

(i) ΔABD≅ΔBAC

(ii) BD=AC

(iii) ∠ABD=∠BAC

Sol.

Given: ABCD is a quadrilateral with AD=BC and ∠DAB=∠CBA.

Proof:

(i) In ΔABD and ΔBAC:

  1. AD = BC (Given)
  2. ∠DAB = ∠CBA (Given)
  3. AB = AB (Common side)
    Therefore, by SAS (Side-Angle-Side) congruence rule, we have ΔABD≅ΔBAC.

(ii) Since ΔABD≅ΔBAC (Proved above), by CPCT, we have BD=AC.

(iii) Since ΔABD≅ΔBAC (Proved above), by CPCT, we have ∠ABD=∠BAC.

3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.

Geometry of Triangle Image 6

Sol.

Given: AD and BC are perpendiculars to line segment AB, and AD=BC.

To prove: CD bisects AB (i.e., OA=OB).

Proof: In ΔOAD and ΔOBC:

  1. AD = BC (Given)
  2. ∠OAD = ∠OBC (Each = 90°, as AD⊥AB and BC⊥AB)
  3. ∠AOD = ∠BOC (Vertically opposite angles)
    Therefore, by AAS (Angle-Angle-Side) congruence rule, ΔOAD≅ΔOBC.
    By CPCT, we have OA=OB.
    Therefore, CD bisects AB.

4. ℓ and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ΔABC≅ΔCDA.

Geometry of Triangles Image 5

Sol.

Given: Line ℓ∥m and line p∥q.

To prove: ΔABC≅ΔCDA.

Proof: In ΔABC and ΔCDA:

  1. Since ℓ∥m and AC is a transversal, ∠BAC = ∠DCA (Pair of alternate angles).
  2. Since p∥q and AC is a transversal, ∠BCA = ∠DAC (Pair of alternate angles).
  3. AC = AC (Common side)
    Therefore, by ASA (Angle-Side-Angle) congruence criteria, ΔABC≅ΔCDA.

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. Show that:

Geometry of Triangles Image 4

(i) ΔAPB≅ΔAQB

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

Sol.

Given: Line ℓ is the bisector of angle A. B is any point on ℓ. BP⊥AP and BQ⊥AQ.

To prove:

(i) ΔAPB≅ΔAQB

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

Proof:

(i) In ΔAPB and ΔAQB:

  1. ∠BAP = ∠BAQ (Since ℓ is the bisector of ∠A)
  2. AB = AB (Common side)
  3. ∠BPA = ∠BQA (Each = 90°, since BP and BQ are perpendiculars)
    Therefore, by AAS (Angle-Angle-Side) congruence rule, ΔAPB≅ΔAQB.

(ii) Since ΔAPB≅ΔAQB (Proved above), by CPCT, BP=BQ.

This means that B is equidistant from the arms of ∠A.

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

Geometry of Triangle Image 3

Sol.

Given: AC=AE, AB=AD, and ∠BAD=∠EAC.

To prove: BC=DE.

Proof: Consider ΔABC and ΔADE.

We need to show ∠BAC = ∠DAE.

We are given ∠BAD = ∠EAC.

Add ∠DAC to both sides of the given equation:

∠BAD + ∠DAC = ∠EAC + ∠DAC

This gives ∠BAC = ∠DAE.

Now, in ΔABC and ΔADE:

  1. AB = AD (Given)
  2. ∠BAC = ∠DAE (Proved above)
  3. AC = AE (Given)
    Therefore, by SAS (Side-Angle-Side) congruence rule, ΔABC≅ΔADE.
    By CPCT, BC=DE.

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. Show that

(i) ΔDAP≅ΔEBP

(ii) AD=BE

Geometry of Triangle Image 2

Sol.

Given: AB is a line segment, P is its mid-point (AP=PB). D and E are points on the same side of AB. ∠BAD=∠ABE and ∠EPA=∠DPB.

To prove:

(i) ΔDAP≅ΔEBP

(ii) AD=BE

Proof:

(i) We are given ∠EPA = ∠DPB.

Add ∠EPD to both sides (Note: Original solution implies adding ∠DPE, which is same as ∠EPD):

∠EPA + ∠EPD = ∠DPB + ∠EPD

This implies ∠APD = ∠BPE.

Now, in ΔDAP and ΔEBP:

  1. AP = PB (Given, since P is the mid-point of AB)
  2. ∠PAD = ∠PBE (Given ∠BAD=∠ABE, and ∠PAD is ∠BAD, ∠PBE is ∠ABE)
  3. ∠APD = ∠BPE (Proved above)
    Therefore, by ASA (Angle-Side-Angle) congruence rule, ΔDAP≅ΔEBP.

(ii) Since ΔDAP≅ΔEBP (Proved above), by CPCT, AD=BE.

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. Show that:

Geometry of Triangles Image 1

(i) ΔAMC≅ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC≅ΔACB

(iv) CM=(1/2)AB

Sol.

Given: ΔABC is right-angled at C (∠BCA=90°). M is the mid-point of AB (AM=BM). CM is produced to D such that DM=CM. D is joined to B.

Proof:

(i) In ΔAMC and ΔBMD:

  1. AM = BM (Given, M is the mid-point of AB)
  2. ∠AMC = ∠BMD (Vertically opposite angles)
  3. CM = DM (Given)
    Therefore, by SAS (Side-Angle-Side) congruence rule, ΔAMC≅ΔBMD.

(ii) Since ΔAMC≅ΔBMD (Proved above), by CPCT, ∠ACM = ∠BDM.

These are alternate interior angles for lines AC and DB with transversal CD. Since alternate interior angles are equal, AC∥DB.

When two parallel lines (AC and DB) are intersected by a transversal BC, the sum of consecutive interior angles is 180°.

∠BCA + ∠DBC = 180°

Given ∠BCA = 90°.

90° + ∠DBC = 180°

∠DBC = 180° − 90°

Therefore, ∠DBC is a right angle (∠DBC = 90°).

(iii) In ΔDBC and ΔACB:

  1. DB = AC (By CPCT from ΔBMD≅ΔAMC)
  2. ∠DBC = ∠ACB (Each = 90°, proved in (ii) and given)
  3. BC = CB (Common side)
    Therefore, by SAS (Side-Angle-Side) congruence rule, ΔDBC≅ΔACB.

(iv) Since ΔDBC≅ΔACB (Proved above), by CPCT, CD = AB.

We are given that DM = CM.

Also, CM + DM = CD.

Substitute DM = CM: CM + CM = CD => 2CM = CD.

Since CD = AB (Proved above), substitute CD = 2CM into CD = AB.

2CM = AB

Therefore, CM=(1/2)AB.

4.0Key Features and Benefits Class 9 Maths Chapter 7 Triangles: Exercise 7.1

  • Clear Statement on SAS Rule: The SAS Rule is shown in a manner to simplify the understanding of triangle congruence.
  • Aligned with the Latest CBSE: Chapter 7 solutions are fully based on future syllabus and guidelines.
  • Step-By-Step Reasoning: Step-by-Step Reasoning allows students to learn the proper way to offer proof.
  • Improves Logical Thinking: Logical reasoning can potentially improve analytical thinking in terms of geometry.
  • Ideal for Preparation in an Exam: An ideal context for developing some understanding of geometric reason

NCERT Class 9 Maths Ch. 7 Triangles Other Exercises:-

Exercise 7.1

Exercise 7.2

Exercise 7.3


NCERT Solutions for Class 9 Maths Other Chapters:-

Chapter 1: Number Systems

Chapter 2: Polynomials

Chapter 3: Coordinate Geometry

Chapter 4: Linear Equations in Two Variables

Chapter 5: Introduction to Euclid’s Geometry

Chapter 6: Lines and Angles

Chapter 7: Triangles

Chapter 8: Quadrilaterals

Chapter 9: Circles

Chapter 10: Heron’s Formula

Chapter 11: Surface Areas and Volumes

Chapter 12: Statistics

Frequently Asked Questions

This exercise primarily focuses on the basic idea of congruence in triangles, especially using the Side-Angle-Side (SAS) congruence rule.

Understanding congruence helps in determining if two figures, particularly triangles, are identical in shape and size, which is crucial for solving geometric problems and proofs.

You can download the reliable and simplified NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.1 in PDF format directly from the provided link.

Yes, working through Exercise 7.1 with its step-by-step reasoning and proof-based questions significantly helps in developing logical thinking and analytical skills in geometry.

Join ALLEN!

(Session 2025 - 26)


Choose class
Choose your goal
Preferred Mode
Choose State
  • About
    • About us
    • Blog
    • News
    • MyExam EduBlogs
    • Privacy policy
    • Public notice
    • Careers
    • Dhoni Inspires NEET Aspirants
    • Dhoni Inspires JEE Aspirants
  • Help & Support
    • Refund policy
    • Transfer policy
    • Terms & Conditions
    • Contact us
  • Popular goals
    • NEET Coaching
    • JEE Coaching
    • 6th to 10th
  • Courses
    • Online Courses
    • Distance Learning
    • Online Test Series
    • International Olympiads Online Course
    • NEET Test Series
    • JEE Test Series
    • JEE Main Test Series
  • Centers
    • Kota
    • Bangalore
    • Indore
    • Delhi
    • More centres
  • Exam information
    • JEE Main
    • JEE Advanced
    • NEET UG
    • CBSE
    • NCERT Solutions
    • Olympiad
    • NEET 2025 Results
    • NEET 2025 Answer Key
    • NEET College Predictor
    • NEET 2025 Counselling

ALLEN Career Institute Pvt. Ltd. © All Rights Reserved.

ISO