NCERT Solutions Class 9 Maths Chapter 7 Triangles : Exercise 7.3

Triangles are important shapes in geometry, and being able to prove their properties will help solve many mathematical problems. In NCERT Solutions Class 9 Maths Chapter 7 Geometry of Triangles equips students to understand and apply theorems effectively.

These NCERT Solutions Class 9 Maths Chapter 7 Triangles Exercise 7.3 is dedicated to proving the properties of isosceles and equilateral triangles, and solving problems based on theorems derived from congruence rules. It emphasizes logical flow and clarity in writing proofs, essential for both exams and higher studies.

1.0Download NCERT Solutions of Class 9 Maths Chapter 7 Triangles Exercise 7.3 : Free PDF

You can improve your geometry skills with simple and easy to comprehend solutions. Download the NCERT Solutions Class 9 Maths Chapter 7 Triangles Exercise 7.3 PDF for offline study, to ease any doubts and help you gain confidence for your exam.

Key Concepts of Exercise 7.3

• Congruent triangles
• The ASA (Angle-Side-Angle) criteria
• The AAS (Angle-Angle-Side) criteria
• The RHS (Right angle-Hypotenuse-Side) criteria
• Geometrical proofs and reasoning
• Application of congruency in triangle based problems

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 7.3 : Detailed Solutions

1. ΔABC and ΔDBC 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) ΔABD≅ΔACD

(ii) ΔABP≅ΔACP

(iii) AP bisects ∠A as well as ∠D

(iv) AP is the perpendicular bisector of BC.

Sol.

Given: ΔABC (with AB=AC) and ΔDBC (with DB=DC) are two isosceles triangles on the same base BC. A and D are on the same side of BC. AP is the line segment extending from A to D, intersecting BC at P.

Proof:

(i) In ΔABD and ΔACD:

1. AB = AC (Given, ΔABC is isosceles)
2. DB = DC (Given, ΔDBC is isosceles)
3. AD = AD (Common side)
   Therefore, by SSS (Side-Side-Side) congruence rule, ΔABD≅ΔACD.

(ii) Since ΔABD≅ΔACD (Proved in (i)), by CPCT, ∠BAD = ∠CAD.

This means AP bisects ∠A.

Now, in ΔABP and ΔACP:

1. AB = AC (Given)
2. ∠BAP = ∠CAP (Proved above, AP bisects ∠A)
3. AP = AP (Common side)
   Therefore, by SAS (Side-Angle-Side) congruence rule, ΔABP≅ΔACP.

(iii) From (ii), we already established that AP bisects ∠A (∠BAP=∠CAP).

Now, to show AP bisects ∠D:

Since ΔABD≅ΔACD (Proved in (i)), by CPCT, ∠BDA = ∠CDA.

Also, in ΔDBP and ΔDCP: (This is generally not the direct way. Let's use angles from big triangles)

From ΔABD≅ΔACD, we have ∠ADB = ∠ADC.

In ΔDBC, since DB=DC, ∠DBC=∠DCB.

Consider ∠ADB and ∠ADC (or ∠BDP and ∠CDP).

From ΔABP≅ΔACP (Proved in (ii)), BP=CP (by CPCT) and ∠APB=∠APC.

In ΔDBP and ΔDCP:

DB = DC (Given)

BP = CP (Proved above)

DP = DP (Common)

Therefore, ΔDBP≅ΔDCP (by SSS congruence).

By CPCT, ∠BDP = ∠CDP.

Thus, AP bisects ∠D.

(iv) From ΔABP≅ΔACP (Proved in (ii)), by CPCT, BP = CP. This means AP bisects BC.

Also, by CPCT, ∠APB = ∠APC.

Since ∠APB and ∠APC form a linear pair, their sum is 180°.

∠APB + ∠APC = 180°

∠APB + ∠APB = 180°

2∠APB = 180°

∠APB = 90°.

Since ∠APB = 90° and AP bisects BC, AP is the perpendicular bisector of BC.

2. AD is an altitude of an isosceles triangle ABC in which AB=AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠A

Sol.

Given: ΔABC is an isosceles triangle with AB=AC. AD is an altitude to BC, meaning AD⊥BC.

To prove:

(i) AD bisects BC (i.e., BD=CD).

(ii) AD bisects ∠A (i.e., ∠BAD=∠CAD).

Proof:

(i) In right-angled ΔADB and right-angled ΔADC:

1. Hypotenuse AB = Hypotenuse AC (Given, ΔABC is isosceles)
2. ∠ADB = ∠ADC (Each = 90°, since AD is an altitude)
3. Side AD = Side AD (Common)
   Therefore, by RHS (Right Angle-Hypotenuse-Side) congruence criteria, ΔADB≅ΔADC.
   By CPCT, BD=CD.
   Thus, AD bisects BC.

(ii) Since ΔADB≅ΔADC (Proved above), by CPCT, ∠BAD=∠CAD.

Thus, AD bisects ∠A.

3. Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of ΔPQR (see figure). Show that:

(i) ΔABM≅ΔPQN

(ii) ΔABC≅ΔPQR

Sol.

Given: AB=PQ, BC=QR, and median AM=median PN. (M is midpoint of BC, N is midpoint of QR).

To prove:

(i) ΔABM≅ΔPQN

(ii) ΔABC≅ΔPQR

Proof:

(i) Given AM is the median of ΔABC, so M is the midpoint of BC. Thus, BM = (1/2)BC.

Given PN is the median of ΔPQR, so N is the midpoint of QR. Thus, QN = (1/2)QR.

We are given BC=QR.

Therefore, (1/2)BC = (1/2)QR, which means BM=QN.

Now, in ΔABM and ΔPQN:

1. AB = PQ (Given)
2. BM = QN (Proved above)
3. AM = PN (Given)
   Therefore, by SSS (Side-Side-Side) congruence criteria, ΔABM≅ΔPQN.

(ii) Since ΔABM≅ΔPQN (Proved above), by CPCT, ∠B = ∠Q.

Now, in ΔABC and ΔPQR:

1. AB = PQ (Given)
2. ∠B = ∠Q (Proved above)
3. BC = QR (Given)
   Therefore, by SAS (Side-Angle-Side) congruence criteria, ΔABC≅ΔPQR.

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

Sol.

Given: In ΔABC, BE⊥AC and CF⊥AB. Also, BE=CF.

To prove: ΔABC is an isosceles triangle (i.e., AB=AC).

Proof: In right-angled ΔBEC and right-angled ΔCFB:

1. Side BE = Side CF (Given altitudes are equal)
2. Hypotenuse BC = Hypotenuse CB (Common side)
3. ∠BEC = ∠CFB (Each = 90°, since BE and CF are altitudes)
   Therefore, by RHS (Right Angle-Hypotenuse-Side) congruence rule, ΔBEC≅ΔCFB.
   By CPCT, ∠EBC = ∠FCB (which means ∠ABC = ∠ACB).
   In ΔABC, since ∠ABC = ∠ACB, the sides opposite to these equal angles are equal.
   Therefore, AB=AC.
   Hence, ΔABC is an isosceles triangle.

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

Sol.

Given: ΔABC is an isosceles triangle with AB=AC. AP⊥BC.

To prove: ∠B=∠C.

Proof: In right-angled ΔAPB and right-angled ΔAPC:

1. Hypotenuse AB = Hypotenuse AC (Given)
2. ∠APB = ∠APC (Each = 90°, since AP⊥BC)
3. Side AP = Side AP (Common side)
   Therefore, by RHS (Right Angle-Hypotenuse-Side) congruence criteria, ΔAPB≅ΔAPC.
   By CPCT, ∠ABP = ∠ACP (which means ∠B=∠C).

4.0Key Features and Benefits: Class 9 Maths Chapter 7 Exercise 7.3

• Thorough download of proofs: we see how to structure and write formal geometrical proofs.
• Concretely defined reasonings for congruency rules: provides clarity to ASA, AAS and RHS through investigations of actual problems. 
• Designed according to CBSE guidelines: created with the new CBSE Board exams front-of-mind. 
• Increase in confidence: provides the opportunity for students to gain confidence with practicing geometry with triangles.  
• Ideal for exams: A useful revision and consolidation activity to master congruency questions for school and board exams.