Let $ABC$ be an acute triangle such that $AB$ is the shortest side of the triangle. Let $D$ be the midpoint of the side $AB$ and $P$ be an interior point of the triangle such that $\angle CAP = \angle CBP = \angle ACB$. Denote by M and $N$ the feet of the perpendiculars from $P$ to $BC$ and $AC$, respectively. Let $p$ be the line through $ M$ parallel to $AC$ and $q$ be the line through $N$ parallel to $BC$. If $p$ and $q$ intersect at $K$ prove that $D$ is the circumcenter of triangle $MNK$.