NCERT Solutions Class 9 Triangles Exercise 7.3

Exercise 7.3 is an important part of Class 9 Triangles, and it strengthens your understanding of triangle congruence rules and their applications. The NCERT Solutions Class 9 Maths Chapter 7 Ex 7.3 provide clear, step-by-step explanations so you can understand how and why each congruence rule is used to justify relationships in a triangle.

This exercise moves beyond basic identification and focuses on applying congruency rules like SSS, SAS, ASA, and AAS to solve reasoning-based questions. It is a scoring exercise and is crucial for the final exams.

What Does Class 9 Triangles Exercise 7.3 Cover?

Class 9 Triangles Exercise 7.3 mainly tests your reasoning using congruence criteria. Questions require you to:

• Identify which sides or angles are equal from a given figure

• Determine the appropriate congruence rule (SSS, SAS, ASA, AAS)

• Prove triangles congruent using proper logical steps

• Deduce additional equal parts after congruence (CPCTC)

With the help of NCERT Solutions for Class 9 Maths Triangles Exercise 7.3, you will learn the correct format for writing congruency proofs. 

Class 9 Maths Chapter 7 Triangles Exercise 7.3 Questions and NCERT Solutions

Class 9 Maths Triangles Exercise 7.3 questions focus on logically proving triangle congruence using given conditions. Each question requires you to carefully examine the diagram, identify matching parts, and apply the correct congruence rule.

Below are the NCERT Solutions for Class 9 Maths Chapter 7 Ex 7.3 that you can use to verify your steps and strengthen your understanding:

Question 1.ABC andDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (See figure). If AD is extended to intersect BC at P, show that:

(i)ABDACD (ii)ABPACP (iii) AP bisectsA as well asD. (iv) AP is the perpendicular bisector of BC.

Solution: (i)ABC is an isosceles triangle.AB = ACDBC is an isosceles triangle.BD = CD Now inABD andACD, AB = AC [Given] BD = CD [Given] AD = AD [Common]ABDACD [By SSS congruency]BAD =CAD [By C.P.C.T.] ……….(i) 

(ii)Now inABP andACP, AB = AC [Given]BAD =CAD [From eq. (i)] AP = APABPACP [By SAS congruency] 

(iii)SinceABPACP [From part (ii)]BAP =CAP [By C.P.C.T.]AP bisectsA. SinceABDACD [From part (i)]ADB =ADC [By C.P.C.T.] ……….(ii) NowADB +BDP =[Linear pair] ……….(iii) AndADC +CDP =[Linear pair] ……….

(iv) From eq. (iii) and (iv),ADB +BDP =ADC +CDPADB +BDP =ADB +CDP [Using (ii)]BDP =CDPDP bisectsD or AP bisectsD. (iv)SinceABPACP [From part (ii)]BP = PC [By C.P.C.T.] ……….

(v) AndAPB =APC [By C.P.C.T.] …….

(vi) NowAPB +APC =[Linear pair]APB +APC =[Using eq. (vi)]2APB =APB =APBC ……….

(vii) From eq. (v), we have BP PC and from (vii), we have proved APB. So, collectively AP is perpendicular bisector of BC. 

Question 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that: (i) AD bisects BC. (ii) AD bisectsA.

 Solution: InABD andACD, AB = AC [Given]ADB =ADC =[ADBC] AD = AD [Common]ABDACD [RHS rule of congruency]BD = DC [By C.P.C.T.]AD bisects BC AlsoBAD =CAD [By C.P.C.T.]AD bisectsA.

Question 3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that (i) ∆ABC ≅∆PQR (ii) ∆ABM ≅∆PQN

Solution:

In ∆ABC, AM is the median. ∴BM =BC ……(1) In ∆PQR, PN is the median. ∴QN =QR …(2) And BC = QR [Given] ⇒BC =QR ⇒BM = QN …(3) [From (1) and (2)] (i) In ∆ABM and ∆PQN, we have AB = PQ , [Given] AM = PN [Given] BM = QN [From (3)] ∴∆ABM ≅∆PQN [By SSS congruency] (ii) Since ∆ABM ≅∆PQN ⇒∠B = ∠Q …(4) [By C.P.C.T.] Now, in ∆ABC and ∆PQR, we have ∠B = ∠Q [From (4)] AB = PQ [Given] BC = QR [Given] ∴∆ABC ≅∆PQR [By SAS congruency]

 Question 4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

Solution:

Since BE ⊥AC [Given]

∴BEC is a right triangle such that ∠BEC = 90° Similarly, ∠CFB = 90° Now, in right ∆BEC and ∆CFB, we have BE = CF [Given] BC = CB [Common hypotenuse] ∠BEC = ∠CFB [Each 90°] ∴∆BEC ≅∆CFB [By RHS congruency] So, ∠BCE = ∠CBF [By C.P.C.T.] or ∠BCA = ∠CBA Now, in ∆ABC, ∠BCA = ∠CBA ⇒AB = AC [Sides opposite to equal angles of a ∆ are equal] ∴ABC is an isosceles triangle. 

Question 5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥BC to show that ∠B = ∠C. 

Solution: We have, AP ⊥BC [Given]

∠APB = 90° and ∠APC = 90° In ∆ABP and ∆ACP, we have ∠APB = ∠APC [Each 90°] AB = AC [Given] AP = AP [Common] ∴∆ABP ≅∆ACP [By RHS congruency] So, ∠B = ∠C [By C.P.C.T.]

How to Prepare Using NCERT Solutions Class 9 Maths Chapter 7 Ex 7.3?

To score full marks in NCERT Solutions Class 9 Maths Chapter 7 Ex 7.3, follow this structured approach:

1. Revise SSS, SAS, ASA, RHS, and AAS thoroughly so you can instantly identify the correct rule while solving proofs.

2. Write each reason clearly when proving triangles congruent. Examiners give marks for each step.

3. Before writing anything, mark equal sides and angles directly on the diagram. This reduces confusion.

4. Once triangles are proved congruent, deducing equal parts using CPCTC is essential. 

If your method is longer or confusing, refer to the ncert solutions for class 9 maths triangles exercise 7.3 to learn efficient steps.