72nd Czech and Slovak Mathematical Olympiad

Let $M$ be the midpoint of $BC$. In point reflection with the centre $M$, denote $L$ the image of $D$ and $J$ the image of $I$. It follows from this symmetry, that $J$ is the incenter of triangle $BCP$ and $L$ is the point where this incircle touches $BC$. Let $KL$ be the diameter of this incircle. Thus, $J$ is the midpoint of the segment $KL$.

It is well known that $D$ is the tangent point of the excircle of triangle $BCP$. This excircle is the image of the incircle in homothety with center $P$ (and with a coefficient greater than $1$). In this homothety is the tangent $BC$ of the excircle the image of the tangent of the incircle, that is parallel to their common tangent $BC$ but has smaller distance from $P$. This tangent, however, passes through the point $K$, since $KL$ is the diameter of the incircle perpendicular to both tangents. Therefore, the homothety maps the point $K$ to the point $D$, and thus the points $D$, $K$, $P$ lie on the same line. This remains to prove that point $E$ also lies on this line. To do this, it suffices to show that the lines $EP$ and $DK$ are parallel. These are, however, the sides of triangles $AEP$ and $LDK$ with the meanlines $IM$ and $MJ$ respectively for which $IM\parallel EP$ and $MJ\parallel DK$; hence the desired relation $EP\parallel DK$ follows, since $M$ is the midpoint of the line segment $IJ$.