First, notice that $MN \parallel BC$, so $\frac{WD}{DB} = \frac{EW}{EC}$. Applying Ceva's theorem to triangle $BCW$ and cevians $WP$, $CD$ and $BE$ we have $\frac{WD}{DB} \cdot \frac{BP}{PC} \cdot \frac{CE}{EW} = 1 \iff BP = PC$, so $WZ$ meets $BC$ at its midpoint $P$.

Now it remains to prove that $P$ lies on line $XY$, which is the radical axis of the two circles. But, because of the reflection and the fact that $MN \parallel BC$,

$\angle MBC = \angle CBW = \angle MDB$, and thus $BC$ is tangent to the circumcircle of $BDM$. The power of $P$ with respect to this circle is $PB^2 = BC^2/4$. Analogously, the power of $P$ with respect to the circumcircle of $CEN$ is $BC^2/4$, and thus $P$ lies on the radical axis of these circles, and we are done.