<Tri-205-2264>, hint

Let <MNA=y, <NMA=x Since MN is tangent to circucircles of ABD (s') and ACD(s")

<MNA(y)=<ADN=<ACN, <NMA(x)= <ADM=<ABM. Let Q be the midpoint of MN then

MQ^2=NQ^2=QA xQD, Line DA cut MN at Q.

<MNC=y+C+A/2, <MBC=B+x, x+y=A/2, <MNC+<MBC=A+B+C=180, M,N,C,B are cyclic

Suppose lines DQ, BM, CN  meet at P, then A,N,P,M are cyclic

Since A,N,P,M are cyclic and PMDN is a parallelogram ( <NPA=<NMA=x=<MDQ)

 (B', C' are midpoints of BD, CD, MN=B'C', PQ=QD), Triangles  PBC and QB'C' are similar , ratio of similitude is 2, PD=2QD, BC=2B'C'

<NQD=<MNP+<NPQ-<C'QD=B+x+x-x=B+x=<QB'D. Hence MN is tangent to the circumcircle of triangle QB'C'. ( Remark : please check my answer)