European Mathematical Cup

Let the point $C$ be the midpoint of the line segment $AB$. We have to prove $MC\parallel KP$. Let us introduce angle $\alpha =\angle BKA$. Notice that $$\begin{array}{rl}\angle BLA& =180^{\circ }-\angle BMA=180^{\circ }-\frac{1}{2}\angle BOA=\\ & =180^{\circ }-\frac{1}{2}(180^{\circ }-\angle BKA)=90^{\circ }+\frac{1}{2}\alpha \end{array}$$ Also, notice that the point $O$ is midpoint of the arc $\text{widearc}AB$. Thus the line $KO$ is bisector of the angle $\angle BKA$. From the two claims above, we deduce that $L$ is incenter of the triangle $ABK$. Moreover, notice that $ML$ is diameter of the circle $k_2$, thus $\angle ABM=90^{\circ }$. Since $BL$ is angle bisector of the angle $\angle ABK$ we deduce that $BM$ is exterior angle bisector of the same angle. Thus, since $M$ lies on angle bisector $KM$ and exterior angle bisector $BM$, $M$ is the center of the excircle for the triangle $ABK$.

Thus, we have to prove that the line passing through the incenter $L$ of the triangle $ABK$ and point of the tangency of incircle of the same triangle is parallel to the line passing through the center of the excircle $M$ and the midpoint $C$ of the line segment $AB$. This is a well known lemma, which completes the proof.