Solution: Let $F=AB \cap CD$ and T be the intersection of EF and $\odot (EAB)$.As you know,T is the Miquel point of complete quadrilateral ABCD.EF(which means that T is the common point of four circles $\odot (EAB)$,$\odot (EDC)$,$\odot (FAD)$ and $\odot (FBC)$ ).Notice 2 cyclic quadrilaterals TEBC and TECD so it's easy to show that $\triangle TAD\sim\triangle TBC$ (a-a).On the other hand,points M,N divide the segment lines AD,BC in the same ratio respectively so we get $\triangle TAM\sim\triangle TBN\Longrightarrow\angle TME=\angle TMA=\angle TNB=\angle TNE\Longrightarrow$ TENM is the cyclic quadrilateral(or T lies on the circle (EMN)) and hence we get $P_{F/ \odot (EMN)}=\overline{FT}\cdot\overline{FE}=\overline{FD}\cdot\overline{FC}=P_{F/ \odot (ABCD)}$.Therefore F lies on the radical axis of $\odot (EMN)$ and $\odot (ABCD)$ so F,X,Y are collinear (which means that AB,XY and CD are concurrent at a point F).We done!