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: Triangle ABC is isosceles in which AB=AC also OB and OC are bisectors of angle B and angle C

To Prove: i) OB = OC ii) AO bisects ∠A

Let's construct a diagram according to the given question.

 

 

i) OB = OC

It is given that in triangle ABC,

AB = AC (given)

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

1/2 ∠ACB = 1/2 ∠ABC

⇒ ∠OCB = ∠OBC (Since OB and OC are the angle bisectors of ∠ABC and ∠ACB)

∴ OB = OC (Sides opposite to equal angles of an isosceles triangle are also equal)

ii) AO bisects ∠A

In ΔOAB and ΔOAC,

AO = AO (Common)

AB = AC (Given)

OB = OC (Proved above)

Therefore,

ΔOAB ≅ ΔOAC (By SSS congruence rule)

Also, we can use an alternative approach as shown below,

∠OBA = ∠OCA (OB and OC bisects angle ∠B and ∠C)

AB = AC (Given)

OB = OC (Proved above)

 ΔOAB ≅ ΔOAC (By SAS congruence rule)

⇒ ∠BAO = ∠CAO (CPCT)

∴ AO bisects ∠A or AO is the angle bisector of ∠A.

☛ Check: NCERT Solutions Class 9 Maths Chapter 7

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Video Solution:

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

NCERT Maths Solutions Class 9 Chapter 7 Exercise 7.2 Question 1

Summary:

If in an isosceles triangle ABC, with AB = AC, and the bisectors of ∠B and ∠C intersect each other at O, we have proved that OB = OC and AO is the angle bisector ∠A. 

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☛ Related Questions:

• In ΔABC,AD is the perpendicular bisector of BC.Show that ΔABC is an isosceles triangle in which AB=AC.
• ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig. 7.31). Show that these altitudes are equal.
• ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig. 7.32). Show that(i) ΔABE ≅ ΔACF(ii) AB = AC, i.e., ABC is an isosceles triangle.
• ABC and DBC are two isosceles triangles on the same base BC (see Fig. 7.33). Show that ∠ABD = ∠ACD.