**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 – Triangles**

**1. In an isosceles triangle ABC, with AB = AC, the bisectors of ****∠**** B and ****∠**** C intersect**** each other at O. Join A to O.**

**Show that :**

**(i) OB = OC**

(ii) AO bisects

(ii) AO bisects

**∠**

**A**

**Solution :**

(i) In triangle ABC, It is given that AB = AC

∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are equal)

(1/2) ∠ACB = (1/2) ∠ABC (OB and OC bisect ∠B and ∠C)

∠OCB = ∠OBC

=> OB = OC Sides opposite to equal angles of a triangle are also equal)**Hence Proved**

(ii) In ∆OAB and ∆OAC,

AB = AC (Given)

AO =AO (Common)

OB = OC (Proved above)

=> ∆OAB ≅ ∆OAC (By SSS congruence rule)

=> ∠BAO = ∠CAO (By CPCT)

=> AO bisects ∠A.**Hence Proved.**

**2. ****In ∆ ABC, AD is the perpendicular bisector of BC (see given figure). Show that ∆ ABC is an isosceles triangle in which AB = AC.**

**Solution :**

In ∆ADC and ∆ADB,

AD = AD (Common)

∠ADC =∠ADB (Each 90˚)

CD = BD (AD is the perpendicular bisector of BC)

=> ∆ADC ≅ ∆ADB (By SAS congruence rule)

=> AB = AC (By CPCT)

=> ABC is an isosceles triangle in which AB = AC.**Hence Proved.**

**3. ****ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see given figure). Show that these altitudes are equal.**

**Solution :**

In ∆AEB and ∆AFC,

∠AEB and ∠AFC (Each 90˚)

∠A = ∠A (Common angle)

AB = AC (Given)

=> ∆AEB ≅ ∆AFC (By AAS congruence rule)

=> BE = CF (By CPCT)**Hence Proved.**

**4. ****ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see given figure). Show that**

**(i) ∆ ABE**

**≅**

**∆ ACF**

**(ii) AB = AC, i.e., ABC is an isosceles triangle.**

** **

**Solution :**

**(i)** In ∆ABE and ∆ACF,

∠AEB and ∠AFC (Each 90˚)

∠A = ∠A (Common angle)

BE = CF (Given)

=> ∆ABE ≅ ∆ACF (By AAS congruence rule)**Hence Proved.**

**(ii)** It has already been proved that ∆ABE ≅ ∆ACF

=> AB = AC (By CPCT)**Hence Proved.**

**5. ABC and DBC are two isosceles triangles on the same base BC (see given figure). Show that ****∠**** ABD = ****∠**** ACD****.**

**Solution :**

Construction :

Let us join AD.

In ∆ABD and ∆ACD,

AB = AC (Given)

BD = CD (Given)

AD = AD (Common side)

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

=> ∠ABD = ∠ACD (By CPCT)**Hence Proved.**

**6. ****∆ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see given figures). Show that ****∠**** BCD is a right angle.**

**Solution :**

In ∆ACD,

AC = AD (Given)

=> ∠ADC = ∠ACD (Angles opposite to equal sides of a triangle are also equal)

In ∆ABC,

AB = AC (Given)

=> ∠ACB = ∠ABC (Angles opposite to equal sides of a triangle are also equal)

By Angle sum property of a triangle, In ∆BCD

∠ABC + ∠BCD + ∠ADC = 180˚

=> ∠ACB + ∠ACB +∠ACD + ∠ACD = 180˚

=> 2(∠ACB + ∠ACD) = 180˚

=> 2(∠BCD) = 180˚

=> ∠BCD = 90˚**Hence Proved.**

**7. ****ABC is a right angled triangle in which ****∠**** A = 90°**** and AB = AC. Find ****∠**** B and ****∠**** C.**

**Solution :**

Given,

AB = AC

=> ∠C = ∠B (Angles opposite to equal sides are also equal)

By Angle sum property of a triangle, In ∆ABC

∠A + ∠B + ∠C = 180˚

=> 90˚ + ∠B + ∠C = 180˚ (∵ ∠C = ∠B)

=> 90˚ + ∠B + ∠B = 180˚

=> 2∠B = 90˚

=> ∠B = 45˚

=> ∠B = ∠C = 45˚**Hence Proved.**

**8. ****Show that the angles of an equilateral triangle are 60° each.**

**Solution :**

Consider an equilateral triangle ABC,

=> AB = BC = AC

=> AB = AC

=> ∠C = ∠B (Angles opposite to equal sides of a triangle are equal)…..1

And also,

AC = BC

=> ∠B = ∠A (Angles opposite to equal sides of a triangle are equal)…..2

From equation 1 and equation 2

∠A = ∠B = ∠C

By Angle Sum property of triangle, In ∆ABC

∠A + ∠B + ∠C = 180°

=> ∠A + ∠A + ∠A = 180°

=> 3∠A = 180°

=> ∠A = 60°

=> ∠A = ∠B = ∠C = 60°

=> In an equilateral triangle, all interior angles are of measure 60º.

**Download NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 – Triangles**

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