In ∆ABC, AB=AC and the bisectors of ∠B and ∠C meet at a point O. prove that BO=CO and the ray AO is the bisector of ∠A.

Given: In ∆ABC, AB=AC and the bisectors of ∠B and ∠C meet at a point O.

To prove: BO=CO and ∠BAO = ∠CAO

Proof:

In , ∆ABC we have,

∠OBC = ∠B

∠OCB = ∠C

But ∠B = ∠C … given

So, ∠OBC = ∠OCB

Since the base angles are equal, sides are equal

∴ OC = OB …(1)

Since OB and OC are bisectors of angles ∠B and ∠C respectively, we have

∠ABO = ∠B

∠ACO = ∠C

∴∠ABO = ∠ACO …(2)

Now in ∆ABO and ∆ACO

AB = AC … given

∠ABO = ∠ACO … from 2

BO = OC … from 1

Thus by SAS property of congruence,

∆ABO ≅ ∆ACO

Hence, we know that, corresponding parts of the congruent triangles are equal

∠BAO = ∠CAO

ie. AO bisects ∠A