# Class 9 NCERT Solutions- Chapter 7 Triangles – Exercise 7.2

### (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]

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]

∴∆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]

### (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,

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

In ∆BDC,

BD=DC

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

∠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           

In ∆ACD,

∠BCD=∠ABC+∠BDC

∠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                    

Also

AC=BC

∠B=∠A                    

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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