Question

in an isosceles triangle ABC the bisector of Angle B and angle C intersect each other at O then o b bisects a true or false​

Answer 1

Answer:

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.

Answer 2

Given,

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

To Find,

prove that 1) OB = OC

                 2) AO bisects ∠A

Solution,

[ firstly draw a triangle according to the given question.]

1) AB = AC            (given)

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

   1/2 ∠ACB = 1/2 ∠ABC

which means, ∠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)

2) AO bisects ∠A

   In ΔOAB and ΔOAC,

   AO = AO   (common)

   AB = AC    ( given)

   OB = OC   ( proved above)

   ∴ ΔOAB ≅ ΔOAC  ( by SSS congruence rule)

  so, ∠BAO = ∠CAO    (CPCT)

  which means, AO bisects ∠A or AO is the angle bisector of ∠A.

Hence, OB = OC and AO bisect ∠A.