Saudi Arabia Mathematical Competitions 2012

Let $O$ be the center of the semicircle and let $R$ be the midpoint of segment $MN$.



In triangle $ANO$ we have $A{N}^2=AR\cdot AO$. Using the power of the point $A$ with respect to the circle we get $$A{M}^2=AP\cdot AE=A{N}^2=AR\cdot AO.(1)$$ From (1) it follows that the quadrilateral $PROE$ is cyclic, hence $RP\perp AE$. Since $FP\perp AE$, we get that the points $F$, $R$, $P$ are collinear.