AoPS - Problem

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

$KO\perp RS$ .