Let $\vartriangle ABC$ be a triangle with orthocenter $H$ and altitudes $AD$, $BE$ and $CF$. Let $D'$, $E' $ and $F'$ be the intersections of the heights $AD$, $BE$ and $CF$, respectively, with the circumcircle of $\vartriangle ABC $, so that they are different points from the vertices of triangle $\vartriangle ABC$. Let $L$, $M$ and $N$ be the midpoints of $BC$, $AC$ and $AB$, respectively. Let $ P$, $Q$ and $R$ be the intersections of the circumcircle with $LH$, $MH$ and $NH$, respectively, such that $ P$ and $ A$ are on opposite sides of $BC$, $Q$ and $A$ are on opposite sides of $AC$ and $R$ and $C$ are on opposite sides of $AB$. Show that there exists a triangle whose sides have the lengths of the segments $D' P$, $E'Q$, and $F'R$.