Let $ABCD$ be a trapezoid of bases $AD$ and $BC$ , with $AD> BC$, whose diagonals are cut at point $E$. Let $P$ and $Q$ be the feet of the perpendicular drawn from $E$ on the sides $AD$ and $BC$, respectively, with $P$ and $Q$ in segments $AD$ and $BC,$ respectively. Let $I$ be the center of the triangle $AED$ and let $K$ be the point of intersection of the lines $AI$ and $CD$. If $AP + AE = BQ + BE$, show that $AI = IK$.