Given :
AB = AC

OB is the bisectors of ∠B
⇒ ∠ABO = ∠OBC = .
OC is the bisectors of ∠C
⇒ ∠ACO = ∠OCB = .
(i) It is given that in triangle ABC, AB = AC
⇒ ∠ACB = ∠ABC
Dividing both sides of equation by 2, we get :
⇒ ∠OCB = ∠OBC
We know that,
Sides opposite to equal angles of a triangle are also equal.
⇒ OB = OC.
Hence, proved that OB = OC.
(ii) In Δ OAB and Δ OAC,
⇒ AO = AO (Common)
⇒ OB = OC (Proved above)
⇒ AB = AC (Proved above)
∴ Δ OAB ≅ Δ OAC (By S.S.S. congruence rule)
We know that,
Corresponding parts of congruent triangles are equal.
⇒ ∠BAO = ∠CAO (By C.P.C.T.)
∴ AO bisects ∠A
Hence, proved that AO bisects ∠A.