Iranian Mathematical Olympiad

Let $L$ be the intersection of $KM$ and $CD$. The quadrilateral $MLCB$ has an incircle centered at $O$. So $OL$ and $OM$ are angle-bisectors of $\angle MLC$ and $\angle BML$, respectively. Further, $BM \parallel CL$, these two arguments yielding $\angle LOM = 90^\circ$. Therefore $LO$ is tangent to the circumcircle of $OKM$, since $\angle MKO = 90^\circ$. Moreover, $\angle LOM = \angle AOD$ implies that
$$\angle LOD = \angle AOM = \angle BAC = \angle LCO.$$

Hence $LO$ is also tangent to the circumcircle of the triangle $DOC$. Therefore we would have
$$LD \cdot LC = LO^2 = LK \cdot LM,$$
so $MKDC$ is cyclic. As desired.