<Qua-2025-788>, hint

OA is the radius of the circumcircle of triangle ABD , BD/sin60 =2OA

Since <BOD =120, <BCD=60, O,B,C,D are concyclic and  so <OCB=<OCD=30

Since <BCK=60 <OCK=30+60=90, OK  is the diameter of the circumcircle of 

triangle OBD. Thus BD/sin120=OK=BD/sin60=2OA.

Hence OK/OA=2OA/OA=2.