jmc

Since $\overline{AB}\parallel \overline{CD}$, we have $\angle AXE=\angle CYX$. Letting $x=\angle AXE$, we have $x=3x-108^{\circ }$. Solving this equation gives $x=54^{\circ }$. We therefore have $\angle BXY=\angle AXE=\boxed{54^{\circ }}$.