Let $R$ and $S$ be the circumcenters of $\bigtriangleup MOQ$ and $\bigtriangleup NOP$, respectively. Note that $\bigtriangleup AOB\cup\{M\}\sim \bigtriangleup DOC\cup\{P\}$ and that $\bigtriangleup BOC\cup\{N\}\sim \bigtriangleup AOD\cup\{Q\}$ so $\angle NOP= \angle NOC+ \angle COP= \angle COQ+ \angle BOM=180^{\circ}-\angle MOQ$ Hence $(MOQ)$ and $(NOP)$ have the same radius. This, and the fact that $MQ=NP$ implies that $\bigtriangleup MRQ\equiv\bigtriangleup NSP$. Now these two triangles must be homothetic but due to $MN\parallel PQ$, we must have $RS\parallel MN\parallel PQ\parallel AC$. The conclusion follows because $KO\perp RS$.