<Tri-2023-2050>, hint

Let M be the midpoint of BC and line OM cut the circumcircle of ABC in A', A". 

Since <A'AA"=90 A'A // l and since line l bisect OA , it also bisects OA'; 2<BA'O=<BAC

2OM=OB, <BMO=90 Hence <BOM=60=<OA'B;

<BAC=2<BA'o=120