<Cir-2025-787>, hint

<AO'B=2X, <AO"B=2Y, then <O'BO"=180-(X+Y)

Let line EO" cut S" at E' and line FO' cut S' at F'.

Since <EE'F=<FF'E=X+Y, E,F, E',F' are concyclic, hence <E'F'F=<E'EF=X. 

Triangles BO'O" and BF'E' are similar

Since MN//EF <BEF=<EBM=X. AE/(sin EBA)=AB/sinX, AE=AB sin(90-Y) /sinX=AB(cosY/sinX)

AB=2R sin X. Hence AE=2R cosY.

<O'BM=<FBN=<BFE=Y, MB=2R cos <O'BM=2R cosY. Thus AE=MB

Similarly AF=BN, Hence MN=AE+AF

<Answer>. <MBE=X=<AEB, <EAB=<EMB, BE is common. Hence triangles AEB and MBE

are congruent . Hence AE=MB