Let $T\in OC$ satisfy $A,B,T,D$ are cyclic Consider that $\angle DBT=\angle DAT=\angle NMT$ So $M,B,T,N$ are cyclic Since $\angle BNC=\angle ABD=\angle ATD$ Hence $N,T,C,D$ are cyclic So $\angle NCD=\angle NTD=\angle ATD-\angle ADN=\angle ABD-\angle MDN=\angle ABM$

[asy][asy] size(9cm); pair A = 2.5*dir(140); pair B = dir(210); pair C = dir(330); pair D = dir(30)*1.5; pair O = extension(A,C,B,D); pair N = extension(C,C+A-B,B,D); pair M = extension(N,N+A-D,A,C); draw(A--B--C--D--A--cycle); draw(B--M--N--C); draw(A--C); draw(B--D); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$O$",O,dir(O)); dot("$M$",M,dir(90)); dot("$N$",N,dir(90)); [/asy][/asy] It suffices to show that $MB\parallel DC$, because $AB\parallel NC$ implies the result. Note that \[\frac{OB}{OD}=\frac{OB}{ON}\cdot\frac{ON}{OD}=\frac{OA}{OC}\cdot\frac{OM}{OA}=\frac{OM}{OC}\]and we are done.

ACGNmath wrote: It suffices to show that $MN\parallel DC$. For correctness, I believe you meant $MB\parallel DC$.