Macedonian Mathematical Olympiad

Due to symmetry reasons, it is enough to show that $KP\parallel MC$ holds.

First we show that the quadrilateral $ACMD$ is inscribed. (1 point) Namely, let us notice that $B$ lies on the line segment $\overline{CD}$, and $A$ and $M$ are on different sides of the line $CD$. From $\angle BDM=\angle DAB$ and $\angle BCM=\angle BAC$, it follows that $\angle DAC=\angle DAB+\angle BAC=\angle BDM+\angle BCM=180^{\circ }-\angle DMC$. (2 points)

Second, we will show that $B$ and $P$ lie on the same arc which passes through points $A$ and $K$. (1 point) For that purpose, we consider two cases:

first case: the point $P$ lies on the segment $BC$; let us notice that points $A$ and $B$ are on the same side of the line $KP$. Let $E$ denote the intersection of the lines $KB$ and $DM$. We have the sequence of equalities $\angle KBP=\angle DBE=\angle BDE=\angle CDM=\angle CAM=\angle KAP$. Then, from $\angle KBP=\angle KAP$ it follows that the quadrilateral $AKPB$ is inscribed. (1 point)

second case: the point $P$ lies on the segment $AC$; this time the points $A$ and $B$ are on different sides of the line $KP$. Again, let $E$ be the intersection of $KB$ and $DM$. We have the following sequence of equalities $180^{\circ }-\angle KBP=\angle DBE=\angle BDE=\angle CDM=\angle CAM=\angle KAP$, from where it follows that the quadrilateral $AKBP$ is inscribed. (1 point)

Therefore we get $\angle APK=\angle ABK=\angle ADB=\angle ADC=\angle AMC$, with which we confirm that $KP\parallel MC$. (2 points)