Junior Balkan Mathematical Olympiad

Let $X'$ be the symmetric point to $Q$ in line $BC$. Now since $\angle CBA = \angle CQM = \angle CX'M$, $\angle BCA = \angle BQM = \angle BX'M$, we have
$$\angle BX'C = \angle BX'M + \angle CX'M = \angle CBA + \angle BCA = 180^\circ - \angle BAC$$
we have that $X' \in \Gamma$. Now since $\angle AX'B = \angle ACB = \angle MX'B$ we have that $A, M, X'$ are collinear. Note that since
$$\angle DCB = \angle DAB = 90^\circ - \angle ABC = \angle OAC = \angle EAC$$
we get that $DBCE$ is an isosceles trapezoid.
Since $BDCT$ is a parallelogram we have $MT=MD$, with $M, D, T$ being collinear, $BD=CT$, and since $BDEC$ is an isosceles trapezoid we have $BD=CE$ and $ME=MD$. Since
$$\angle BTC = \angle BDC = \angle BED, \quad CE = BD = CT \text{ and}$$

$ME=MT$ we have that $E$ and $T$ are symmetric with respect to the line $BC$. Now since $Q$ and $X'$ are symmetric with respect to the line $BC$ as well, this means that $QX'ET$ is an isosceles trapezoid which means that $Q, X', E, T$ are concyclic. Since $X' \in \Gamma$ this means that $X \equiv X'$ and therefore $A, M, X$ are collinear.

Denote by $H$ the orthocenter of $\triangle ABC$. We use the following well known properties:
(i) Point $D$ is the symmetric point of $H$ with respect to $BC$. Indeed, if $H_1$ is the symmetric point of $H$ with respect to $BC$ then $\angle BH_1C + \angle BAC = 180^\circ$ and therefore $H_1 \equiv D$.
(ii) The symmetric point of $H$ with respect to $M$ is the point $E$. Indeed, if $H_2$ is the symmetric point of $H$ with respect to $M$ then $BH_2CH$ is a parallelogram, $\angle BH_2C + \angle BAC = 180^\circ$ and since $EB \parallel CH$ we have $\angle EBA = 90^\circ$.
Since $DETH$ is a parallelogram and $MH=MD$ we have that $DETH$ is a rectangle. Therefore $MT=ME$ and $TE \perp BC$ implying that $T$ and $E$ are symmetric with respect to $BC$. Denote by $Q'$ the symmetric point of $Q$ with respect to $BC$. Then $Q'ETQ$ is an isosceles trapezoid, so $Q'$ is a point on the circumscribed circle of $\triangle ETQ$. Moreover $\angle BQ'C + \angle BAC = 180^\circ$ and we conclude that $Q' \in \Gamma$. Therefore $Q' \equiv X$.
It remains to observe that $\angle CXM = \angle CQM = \angle CBA$ and $\angle CXA = \angle CBA$ and we infer that $X, M$ and $A$ are collinear.