<Tri-2025-2255>, hint

Since B,C,A,D are concyclic on S', <ADB=180-A,

AB cut circle O'O"A at E,   let M be the midpoint of AC. Since O'O"AD are concyclic 

<ADO'=<AO"M=90-A/2, Since A, D, E, O' are concyclic  <O'DE=<O'AB=90-A. 

Hence <BDE+EDO'+O'DA=<ADB ---> <BDE+90-A+90-A/2=180-A   <BDE=A/2

Since <DEA=DO'A=<A. In triangle ADE  <DAE+<DEA+<ADE(=<ADB-<EDB)(=180-A-A/2)=180--> 

<DAE+A+180-A-A/2=180. <DAE=A/2=<BDE