Question 179
Let $P$ be any point of the angle bisector $A$ in triangle $ABC$, and let $A'$, $B'$ be, $C'$ respective points of the straights $BC$, $CA$, $AB$, such that $PA'$ is perpendicular to $BC$, $PB'$ is perpendicular to $CA$ and $PC'$ is perpendicular to $AB$. It shows that $PA'$ and $B'C'$ are cut on the median $AM$, $M$ being the midpoint of $BC$.