In the given figu

Given: AB = AC

Construction: join OA, OB and OC

Proof:

Consider ΔAOB and ΔAOC

Here,

OC = OB (radius)

OA = OA (common)

AB = AC (given)

∴ By SSS congruency

ΔAOB ΔAOC

∴ ∠OAC = ∠OAB (by C.P.C.T)

Hence, we can say that OA is the bisector of ∠BAC, that is O lies on the bisector of ∠BAC.