Let $X^{\prime}$ be the symmetric point to $Q$ in line $B C$. Now since $\angle C B A=\angle C Q M=\angle C X^{\prime} M$, $\angle B C A=\angle B Q M=\angle B X^{\prime} M$, we have

$$\angle B X^{\prime} C=\angle B X^{\prime} M+\angle C X^{\prime} M=\angle C B A+\angle B C A=180^{\circ}-\angle B A C$$

we have that $X^{\prime} \in \Gamma$. Now since $\angle A X^{\prime} B=\angle A C B=\angle M X^{\prime} B$ we have that $A, M, X^{\prime}$ are collinear. Note that since

$$\angle D C B=\angle D A B=90^{\circ}-\angle A B C=\angle O A C=\angle E A C$$

we get that $D B C E$ is an isosceles trapezoid.

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Since $B D C T$ is a parallelogram we have $M T=M D$, with $M, D, T$ being collinear, $B D=C T$, and since $B D E C$ is an isosceles trapezoid we have $B D=C E$ and $M E=M D$. Since

$$\angle B T C=\angle B D C=\angle B E D, \quad C E=B D=C T \quad \text { and } \quad M E=M T$$

we have that $E$ and $T$ are symmetric with respect to the line $B C$. Now since $Q$ and $X^{\prime}$ are symmetric with respect to the line $B C$ as well, this means that $Q X^{\prime} E T$ is an isosceles trapezoid which means that $Q, X^{\prime}, E, T$ are concyclic. Since $X^{\prime} \in \Gamma$ this means that $X \equiv X^{\prime}$ and therefore $A, M, X$ are collinear.

Alternative solution. Denote by $H$ the orthocenter of $\triangle A B C$. We use the following well known properties:
(i) Point $D$ is the symmetric point of $H$ with respect to $B C$. Indeed, if $H_{1}$ is the symmetric point of $H$ with respect to $B C$ then $\angle B H_{1} C+\angle B A C=180^{\circ}$ 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 $B H_{2} C H$ is parallelogram, $\angle B H_{2} C+\angle B A C=180^{\circ}$ and since $E B \| C H$ we have $\angle E B A=90^{\circ}$.

Since $D E T H$ is a parallelogram and $M H=M D$ we have that $D E T H$ is a rectangle. Therefore $M T=M E$ and $T E \perp B C$ implying that $T$ and $E$ are symmetric with respect to $B C$. Denote by $Q^{\prime}$ the symmetric point of $Q$ with respect to $B C$. Then $Q^{\prime} E T Q$ is isosceles trapezoid, so $Q^{\prime}$ is a point on the circumcircle of $\triangle E T Q$. Moreover $\angle B Q^{\prime} C+\angle B A C=180^{\circ}$ and we conclude that $Q^{\prime} \in \Gamma$. Therefore $Q^{\prime} \equiv X$.

It remains to observe that $\angle C X M=\angle C Q M=\angle C B A$ and $\angle C X A=\angle C B A$ and we infer that $X, M$ and $A$ are collinear.