**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.3 – Triangles****. This Exercise contains 5 questions, for which detailed answers have been provided in this note. In case you are looking at studying the remaining Exercise for Class 9 for Maths NCERT solutions for Chapter 7 or other Chapters, you can click the link at the end of this Note.**

**1. ∆ ABC and ∆ DBC 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) ∆ ABD ****≅**** ∆ ACD
**

**(ii) ∆ ABP**

**≅**

**∆ ACP**

**(iii) AP bisects**

**∠**

**A as well as**

**∠**

**D.**

**(iv) AP is the perpendicular bisector of BC.**

** **

**Solution :**

**(i)** In ∆ABD and ∆ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common)

=> ∆ABD ≅ ∆ACD (By SSS congruence rule)

=> ∠BAD = ∠CAD (By CPCT)

=> ∠BAP = ∠CAP ..(1)

**(ii)** In ∆ABP and ∆ACP,

AP = AP (Common)

∠BAP = ∠CAP [From equation (1)]

AB = AC (Given)

=> ∆ABP ≅ ∆ACP (By SAS congruence rule)

=> BP = CP (By CPCT) ..(2)

**(iii)** From Equation (1),

∠BAP = ∠CAP

=> AP bisects ∠A.

In ∆BDP and ∆CDP,

BD = CD (Given)

DP = DP (Common)

BP = CP [From equation (2)]

=> ∆BDP ≅ ∆CDP (By SSS Congruence rule)

=> ∠BDP = ∠CDP (By CPCT) ..(3)

=> AP bisects ∠D.

(iv) ∆BDP ≅ ∆CDP

=> ∠BPD = ∠CPD (By CPCT) ..(4)

=> ∠BPD + ∠CPD = 180˚ (Since, BPC is a straight line)

=> ∠BPD + ∠BPD = 180˚

=> 2∠BPD = 180˚ [From Equation (4)]

=> ∠BPD = 90˚ …(5)

From Equations (2) and (5),

We can say that,

AP is the perpendicular bisector of BC.

**Hence Proved.**

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

**(i) AD bisects BC**

(ii) AD bisects

(ii) AD bisects

**∠**

**A**

**Solution :**

**(i)** In ∆BAD and ∆CAD,

∠ADB = ∠ADC (Each 90˚ as AD is an altitude)

AD = AD (Common)

AB = AC (Given)

=> ∆BAD ≅ ∆CAD (By RHS Congruence rule)

=> BD = CD (By CPCT)

=> AD bisects BC.

**(ii)** And also,

∠BAD = ∠CAD [By CPCT]

=> AD bisects ∠A.

**Hence Proved.**

**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 ∆ PQR (see Figure). Show that:
**

**(i) ∆ ABM**

**≅**

**∆ PQN**

(ii) ∆ ABC

(ii) ∆ ABC

**≅**

**∆ PQ**

**R**

**Solution :**

**(i)** In ∆ABC,

AM is the median to BC.

= > BM = (1/2) BC

In ∆PQR,

PN is the median to QR.

=> QN = (1/2) QR

We know that,

BC = QR

=> (1/2) BC = (1/2) QR

=> BM = QN ..(1)

In ∆ABM and ∆PQN,

AB = PQ (Given)

AM = PN (Given)

BM = QN [From Equation (1)]

∆ABM ≅ ∆PQN (By SSS congruence rule)

∠ABM = ∠PQN (By CPCT)

∠ABC = ∠PQR ..(2)

(ii) In ∆ABC and ∆PQR,

AB = PQ (Given)

∠ABC = ∠PQR [From Equation (2)]

BC = QR (Given)

=> ∆ABC ≅ ∆PQR (By SAS congruence rule)

**Hence Proved.**

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

**Solution :**

In ∆BEC and ∆CFB,

∠BEC = ∠CFB (Each 90°)

BC = CB (Common)

BE = CF (Given)

=> ∆BEC ≅ ∆CFB (By RHS congruency)

=> ∠BCE = ∠CBF (By CPCT)

=> AB = AC (Sides opposite to equal angles of a triangle are equal)

=> ∆ABC is isosceles.

**Hence Proved.**** **

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

**Solution :**

In ∆APB and ∆APC,

∠APB = ∠APC (Each 90˚)

AB =AC (Given)

AP = AP (Common)

=> ∆APB ≅ ∆APC (Using RHS congruence rule)

=> ∠B = ∠C (By using CPCT)

**Hence Proved.**

**NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.3 – Triangles, has been designed by the NCERT to test the knowledge of the student on the following topics:-**

**Some More Criteria for Congruence of Triangles**

– SSS congruence rule

– RHS congruence rule

**The next Exercise for** **NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.4 – Triangles can be accessed by clicking here.**

**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.3 – Triangles**

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