72nd Czech and Slovak Mathematical Olympiad

Obviously, $MP$ intersects the median $BN$ between points $B$ and $G$, so the point $K$ lies on the ray $PM$ and $BKGP$ is cyclic. Similarly, the point $L$ lies on the ray $NM$ and $CLGN$ is cyclic. Due to $MP\parallel CA$ and $MN\parallel BA$ we have $$|\angle BPK|=|\angle BPM|=|\angle BAC|=|\angle MNC|=|\angle LNC|,$$ while the two cyclic quadrilaterals imply $$|\angle BKP|=|\angle BGP|=|\angle NGC|=|\angle NLC|.$$

We see that triangles BPK and CNL are similar according to the condition AA. By the condition SAS also triangles ABK and ACL are similar since (i) $|\angle ABK|=|\angle PBK|=|\angle NCL|=|\angle ACL|$, (ii) $\frac{|AB|}{|BK|}=2\cdot \frac{|PB|}{|BK|}=2\cdot \frac{|NC|}{|CL|}=\frac{|AC|}{|CL|}.$ Thus, the equality $|\angle BAK|=|\angle CAL|$ is proved.