We are going to demontrate first that AB=BC showing that triangles ABD and BCD have identical measurements.

AD=CD (radius of circle), also BD is a common side of both ABD and BCD triangles and we know that these triangles are right angle triangles so using pythagora's theorem we demonstrate that AB=BC

Actually by pythagora's theorem in ABD triangle AB^2=BD^2-AD^2=BD^2-CD^2 or by pythagora's theorem in BCD BD^2-CD^2=BC^2 so we get AB^2=BC^2 and AB=BC