2003 OIM Problems/Problem 5

Problem

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