Answer 1

It is given that,∠ABC is on circumference of circle BD is passing through centre.  

Construction: Join A and C to form AC and extend BO to D such that BD be the perpendicular bisector of AC.

Now in  \[\bigtriangleup BDA \text{ and }  \bigtriangleup BDC\] we have

AD = CD           (BD is the perpendicular bisector) 

So ,

\[\angle BDA = \angle BDC = 90° \]

BD = BD  (Common)

\[\bigtriangleup BDA \cong \bigtriangleup BDC \left( \text{ SAS congruency criterion }  \right)\]

Hence AB =BC   (by cpct)