Class 9 NCERT Solutions- Chapter 7 Triangles - Exercise 7.2

In Chapter 7 of the Class 9 NCERT textbook, you will explore the properties and theorems related to the triangles. Exercise 7.2 focuses on proving congruence in the triangles using the various congruence rules. Understanding the fundamentals of triangles is essential for solving problems related to geometry. Let's begin by discussing what a triangle is and the key concepts involved in this chapter.

What is a Triangle?

A triangle is a three-sided polygon with three vertices and three edges. It is one of the basic shapes in geometry. The sum of the interior angles of the triangle is always 180°. Based on the sides and angles triangles can be classified into different types such as:

• Equilateral triangle: All sides and angles are equal.
• Isosceles triangle: Two sides and two angles are equal.
• Scalene triangle: All sides and angles are different.

Question 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 ∠A

Solution:

Given: (i) An isosceles ∆ABC in which AB=AC

            (ii) bisects of ∠B and ∠C each other at O.

Show: (i) OB=OC

          (ii) AO bisects ∠A (∠1=∠2)

(i) In ∆ABC,

AB = AC

∠B =∠C                          [angles opposite to equal sides are equal]

1/2 ∠B = 1/2∠C 

∠OBC=∠OCB

∴OB = OC                      [sides opposite equal ∠ are equal]

(ii) In ∆AOB and ADC 

AB = AC                         [given side] 
1/2 ∠B = 1/2∠C 
∠ABO = ∠ACO               [Angle]
BO = OC                        [proved above side]
∴∆AOB ≅ AOC 
Thus ∠1 = ∠2
Therefore, AO bisects ∠A

Question 2. In ΔABC, AD is the perpendicular bisector of BC (see fig.). Show that ΔABC is an isosceles triangle in which AB = AC.

Solution:

Given: AD is ⊥ bisector of BC

Show: AB=BC

In ∆ABD and ∆ACD

BD=DC                    [AD is ⊥ bisector side] 

∠ADB=∠ADC           [Each 90° angle] 

AD=AD                    [common side] 

∴∆ABD≅∆ACD        [S.A.S]

AB=AC                     [C.P.C.T]

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

Solution:

Given: AB=AC,BE and CF are altitudes 

Show: BE=CF

In ∆AEB and ∆AFC,

∠E=∠F                     [Each 90° angle] 

∠A=∠A                    [common angle] 

AB=AC                    [given side] 

∴∆AEB≅∆AFC        [A.A.S]

BE=CF                    [C.P.C.T]

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

(i) ΔABE ≅ ΔACF

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

Solution:

Given: Altitudes BE and CF to sides AC and AB are equal

Show: (i) ΔABE ≅ ΔACF

             (ii) AB = AC

(i) In ∆ABF and ∆ACF,

∠E=∠F                    [Each 90° angle] 

∠A=∠A                   [common angle] 

AB=AC                   [given] S

∴∆AEB≅∆AFC       [A.A.S]

(ii) AB=AC             [C.P.C.T]

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

Solution:

Given: AB=AC,BD=DC

Show: ∠ABD = ∠ACD

In ∆ABD,

AD=AC

∴∠1=∠2                  [angle opposite to equal sides are equal] [1]

In ∆BDC,

BD=DC

∴∠3=∠4                  [angle opposite to equal sides are equal] [2]

Adding 1 and 2

∠1+∠2= ∠2+∠4

∠ABD=∠ACD

Question 6. ΔABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see Fig.). Show that ∠BCD is a right angle.

Solution:

In ∆ABC,

AB=AC

∠ACB=∠ABC           [1]

In ∆ACD,

AC=AD

∠ACD=∠ADC          [2]

Adding 1 and 2

∠ACB+∠ACD=∠ABC+∠ADC 

∠BCD=∠ABC+∠BDC

Adding ∠BCD on both side

∠BCD+∠BCD=∠ABC+∠BDC+∠BCD

2∠BCD=180°

∠BCD=(180°)/2=90°

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

Solution:

Find: ∠B=? and ∠C?

In ∆ABC,

AB=AC

∴∠B=∠C                           [angle opposite to equal side are equal]

∠A+∠B+∠C=180°            [angle sum property of triangle]

90°+∠B+∠B=180°

2∠B=180°-90°

∠B=(90°)/2=45°

Therefore, ∠B=45° and ∠C =45°

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

Solution:

Given: Let ∆ABC is an equilateral ∆ 

Show: ∠A=∠B=∠C=60°

In ∆ABC,

AB=AC

∠B=∠C                    [1]

Also

AC=BC

∠B=∠A                    [2]

From 1,2 and 2

∠A=∠B=∠C

In ∆ABC,

∠A+∠B+∠C=180°      [angle sum property of triangle]

∠A+∠A+∠A=180°

3∠A=180°

∠A=(180°)/3=60°

∠A=60° 

∴∠B=60° and ∠C=60°

Conclusion

Exercise 7.2 in Chapter 7 provides an opportunity to apply congruence rules such as the SAS, ASA and SSS to prove the equality of the triangles. These rules are essential tools in geometry for the establishing the relationship between the triangles and their properties. By mastering these concepts students can strengthen their understanding of the geometric principles.