51st Ukrainian National Mathematical Olympiad, 4th Round

Drop the perpendicular $CL$ to the line $DM$. Then in the triangle $MCL$ we know: $\angle MLC=90^{\circ }$, $\angle LMC=45^{\circ }$. Therefore, $\angle LCM=45^{\circ }$ and $CL=\frac{1}{\sqrt{2}}CM$ (fig. 36).

Since $\angle LCD=60^{\circ }$, from the right triangle $LCD$ we find that $CD=2CL=\sqrt{2}CM$. So, $AB=CD=\sqrt{2}CM=\sqrt{2}BM$, and since $\angle ABM=45^{\circ }$, we obtain that $\angle BMA=90^{\circ }$.

Then $MK$ is a median from the vertex of the right angle in the right triangle $AMB$, and so $KM=BK=AK$. Therefore, the quadrilateral $KBCM$ is a deltoid, its diagonals are perpendicular and intersect at the point $O$.

Thus in the triangle $BOK$ we have that $\angle BOK=90^{\circ }$ and $\angle OBK=\angle ABC-\angle MBC=105^{\circ }-60^{\circ }=45^{\circ }$. This implies that $\angle BKO=45^{\circ }$.