Question 37
In the square $ABCD$, let $P$ and $Q$ be points belonging to the sides $BC$ and $CD$ respectively, different from the ends, such that $BP = CQ$. $X$ and $Y$, $X \neq Y$ are considered points, belonging to the $AP$ and $AQ$ segments respectively. Show that, whatever $X$ and $Y$, there exists a triangle whose sides have the lengths of the segments $BX$, $XY$ and $DY$.