A nice problem. The given condition is equivalent to $\angle BMC=\angle AMN\iff AMN,BMC$ are similar, iff $\frac{AM}{AN}=\frac{BM}{2BN}$. This is equivalent to $\frac{AN}{BN}=\frac{2AM}{BM}$, which, in turn, is equivalent to what we want to prove ($AND,BNM$ are similar).

Oh. Now I really don't understand why my proof took me 1 page. Anyway, the condition that the circumcircle of triangle BMC touches the line AB and the condition that the circumcircle of triangle AND touches the line AB are both equivalent to the condition that the quadrilateral AMNB is harmonic. Darij

Yes, this follows easily: we have $AN\cdot BM=2AM\cdot BN$, and if we also take Ptolemy's Theorem into consideration, we find that the products of the opposite sides are equal for the quadrilateral $ABMN$.

Let $ABCD$ be a convex quadrilateral, and let $M$ and $N$ be the midpoints of its sides $AD$ and $BC$, respectively. Assume that the points $A, B, M, N$ are concyclic, and the circumcircle of triangle $BMC $ touches the line $AB$. Show that the circumcircle of triangle $AND$ touches the line $AB$.