Question

O is the centre of the circle, bo is the bisector of angle ABC. show that AB=AC..Please please urgent

Answer 1

Draw line OD ⊥ AB and line OE ⊥ BC. So now we have the following:
∠ABO = ∠CBO - given as OB is bisector of ∠ABC (A)
∠BDO = ∠BEO = 90°  - by construction(A)
side BO = BO = common side ----- given(A)

Hence Δ DBO ≡ Δ EBO -- by AAS postulate
∴ OD = OE  ---------- by CPCTC 
∴ AB = BC  --- Chords that are equidistant from the center of the circle are equal(or congruent)

Answer 2

bo is the bisector of angleABC
angleABO=angleCBO
if we draw perpendiculars to side AB and BC from point O
let OP and OQ be perpendiculars to side AB and BC respectively
then in triangles OBP and OBQ
anglePBO=angleQBO (given)
angleOPB=angleOQB=90
sideOB=OB
then triangles OBP and OBQ are congruent (AAS property)
then OP=OQ   by c.p.c.t.e
so AB=BC   chords equidistant from the centre of the circle are equal