<Qua-2025-755>, hint

Let OA=a, OB=b, OC=c, OD=d

Since <BMD =90, OM^2 =bxd Similarly ON^2=axc.

Triangles OAB and OCD are similar , so a/b=c/d

OM^2/ON^2=bd/ac in triangle OMN, Triangle OBA (b/a)^2=(b/a) x (d/c)

Thus Triangles OMN and OBA are similar.

In triangle OMN E is the midpoint of MN and <MON=90 hence EM=EO=EN

<AEO=<ABO+<EOB(=<BAO=<MNO)=90