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
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.
☛ 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.