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.

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.

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?

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:

(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.

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.

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:

(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.

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

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:

(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