(1) Note that $a$ is the radical axis of $\Omega$ and $\Omega_1$, $x$ is the radical axis of $\Omega$ and $\omega$, and $B_0C_0$ is the radical axis of $\Omega_1$ and $\omega$. Based on the fact that the three radical axes determined by three circles are concurrent, we obtain: the three lines $a, x$ and $B_0C_0$ are concurrent.

(2) Let $O$ be the circumcenter of $\triangle ABC$, $A_0$ be the midpoint of $BC$, and $Q$ be the foot of the perpendicular from $A_0$ to $B_0C_0$. Note that $\angle WAO = \angle WQO = \angle WXO = 90^\circ$, so the five points $A, W, X, O, Q$ are concyclic. Furthermore, note that reflection about $B_0C_0$ maps $A$ to $D$, and reflection about $OW$ maps $A$ to $X$. Therefore,
$$\angle WQD = \angle WQA = \angle WXA = \angle WAX = \angle WQX.$$
Therefore the three points $Q, D, X$ are collinear.
Finally, note that the spiral similarity (homothety-rotation) centered at the centroid $G$, with ratio $1:2$ and rotation angle $180^\circ$, maps $\triangle ABC$ to $\triangle A_0B_0C_0$, and simultaneously maps $AD$ to $A_0Q$.
Hence the three points $D, G, Q$ are collinear, and thus the three points $D, G, X$ are collinear. This completes the proof.