The problem is not clear to me. Should not the secant MN pass through the intersection H of the tangent t with the center line $O_1O_2$ and should not the points N, D, C be collinear? Assuming this is so, let the secant MN meet the circle $S_2$ again at P and let the normal to AM through B meet the circle $S_2$ again at Q. Because of similarity of the circles $S_1, S_2$ with the similarity center H, $BP \parallel AM$, hence, $BP \perp BQ$, so that PQ is another diameter of the circle $S_2$. The diameters BC, PQ of $S_2$ meet at its center $O_2$. Consider the cyclic pentagon BPCQN as a degenerate cyclic hexagon B'BPCQN with $B' \equiv B$, so that $BB' \equiv t$. By Pascal's theorem, the intersections $H = BB' \cap NP \equiv t \cap MN,$ $O_2 \equiv BC \cap PQ,$ $D' \equiv NC \cap B'Q \equiv NC \cap BQ$ are collinear, i.e., $D' \in HO_2$. But since $HO_2 \equiv O_1O_2,$ we have $D' \equiv O_1O_2 \cap BQ \equiv D,$ the points $D \equiv D'$ are identical and N, D, C are collinear.

Attachments:

I read the original problem and it states that $M$ and $N$ are the intersection points of $S_{1}$ and $S_{2}$.

Assuming M, N are intersections of the circles $\mathcal S_{1},\ \mathcal S_{2},$ let $H \equiv AB \cap O_{1}O_{2}$ be their external similarity center. Let a parallel to AM through B meet $\mathcal S_{2}$ again at P, PM passes through H. Let the perpendicular from B to AM meet $\mathcal S_{2}$ again at Q. $\angle PBQ$ is right, PQ is a diameter of $\mathcal S_{2}.$ By Pascal's theorem for the degenerate hexagon BBPCQM, the intersections $H \equiv BB \cap PM,$ $O_{2}\equiv BC \cap PQ,$ $D \equiv BQ \cap CM$ are collinear, $D \in HO_{2}\equiv O_{1}O_{2}.$