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Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.2
  • Last Updated : 13 Jan, 2021

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°

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