Macedonian Mathematical Competitions

From the conditions in the problem we have $\overline{AM} = \overline{MB}$, $CM \perp AB$, $MN \perp BC$, $\overline{MS} = \overline{SN}$. Let $P$ be a point on $BC$ such that $MP \parallel AN$. From $\triangle ANB$ we have $\overline{AM} = \overline{MB}$ and $MP \parallel AN$, which implies that $MP$ is a median in $\triangle ANB$ and $\overline{NP} = \overline{PB}$. From $\triangle MBN$ we have $\overline{MS} = \overline{SN}$ and $\overline{NP} = \overline{PB}$ which implies that $SP$ is a median in $\triangle MBN$ and $SP \parallel MB$. Let $Q$ be the intersection point of the lines $SP$ and $CM$. Because $CM \perp AB$ and $PQ \parallel AB$, we have $PQ \perp CM$. In the triangle $\triangle MPC$ $MN \perp PC$, $PQ \perp CM$ and $\{S\} = MN \cap PQ$. Hence $S$ is an orthocenter in $\triangle MPC$. Hence $CS \perp MP$. $MP \parallel AN$ implies $CS \perp AN$.