, 2011

If $AB = AC$, then the statement is trivial. So without loss of generality we may assume $AB < AC$. Denote the tangents to $\Omega$ at points $A$ and $X$ by $a$ and $x$, respectively.
Let $\Omega_{1}$ be the circumcircle of triangle $AB_{0}C_{0}$. The circles $\Omega$ and $\Omega_{1}$ are homothetic with center $A$, so they are tangent at $A$, and $a$ is their radical axis. Now, the lines $a$, $x$, and $B_{0}C_{0}$ are the three radical axes of the circles $\Omega$, $\Omega_{1}$, and $\omega$. Since $a \nmid\nmid B_{0}C_{0}$, these three lines are concurrent at some point $W$.
The points $A$ and $D$ are symmetric with respect to the line $B_{0}C_{0}$; hence $WX = WA = WD$. This means that $W$ is the center of the circumcircle $\gamma$ of triangle $ADX$. Moreover, we have $\angle WAO = \angle WXO = 90^{\circ}$, where $O$ denotes the center of $\Omega$. Hence $\angle AWX + \angle AOX = 180^{\circ}$.

Denote by $T$ the second intersection point of $\Omega$ and the line $DX$. Note that $O$ belongs to $\Omega_{1}$. Using the circles $\gamma$ and $\Omega$, we find
$$\angle DAT = \angle ADX - \angle ATD = \frac{1}{2}\left(360^{\circ} - \angle AWX\right) - \frac{1}{2} \angle AOX = 180^{\circ} - \frac{1}{2}(\angle AWX + \angle AOX) = 90^{\circ}.$$
So, $AD \perp AT$, and hence $AT \parallel BC$. Thus, $ATCB$ is an isosceles trapezoid inscribed in $\Omega$.
Denote by $A_{0}$ the midpoint of $BC$, and consider the image of $ATCB$ under the homothety $h$ with center $G$ and factor $-\frac{1}{2}$. We have $h(A) = A_{0}$, $h(B) = B_{0}$, and $h(C) = C_{0}$. From the symmetry about $B_{0}C_{0}$, we have $\angle TCB = \angle CBA = \angle B_{0}C_{0}A = \angle DC_{0}B_{0}$. Using $AT \parallel DA_{0}$, we conclude $h(T) = D$. Hence the points $D$, $G$, and $T$ are collinear, and $X$ lies on the same line.

We define the points $A_{0}$, $O$, and $W$ as in the previous solution and we concentrate on the case $AB < AC$. Let $Q$ be the perpendicular projection of $A_{0}$ on $B_{0}C_{0}$.
Since $\angle WAO = \angle WQO = \angle OXW = 90^{\circ}$, the five points $A$, $W$, $X$, $O$, and $Q$ lie on a common circle. Furthermore, the reflections with respect to $B_{0}C_{0}$ and $OW$ map $A$ to $D$ and $X$, respectively. For these reasons, we have
$$\angle WQD = \angle AQW = \angle AXW = \angle WAX = \angle WQX.$$
Thus the three points $Q$, $D$, and $X$ lie on a common line, say $\ell$.

To complete the argument, we note that the homothety centered at $G$ sending the triangle $ABC$ to the triangle $A_{0}B_{0}C_{0}$ maps the altitude $AD$ to the altitude $A_{0}Q$. Therefore it maps $D$ to $Q$, so the points $D$, $G$, and $Q$ are collinear. Hence $G$ lies on $\ell$ as well.