Solution:

Since by hypothesis $\angle MXA = \angle DAC = \angle MAB$ holds, the two triangles $MXA$ and $MAB$ are similar. Hence
$$\frac{XM}{AM} = \frac{AM}{BM}$$
Let $H$ be the foot of the perpendicular drawn from $A$ to $BC$. From now on we assume that $H$ lies between $B$ and $D$. The other case, namely $D$ between $B$ and $H$, is treated analogously. Since $M$ is the midpoint of the hypotenuse in the right triangle $AHD$, we have $AM = HM$. Hence the previous equality of ratios becomes
$$\frac{XM}{HM} = \frac{HM}{BM}$$
from which we deduce the similarity of the two triangles $XMH$ and $HMB$. From this similarity follows the equality $\angle HXM = \angle BHM$, but $\angle BHM = 180^\circ - \angle MHD = 180^\circ - \angle MDH = \beta + \frac{\alpha}{2}$, where in the second equality we use that the triangle $MHD$ is isosceles, since $AHD$ is a right triangle and $M$ is the midpoint of the hypotenuse, and in the third we use that the sum of the angles of the triangle $ABD$ is $180^\circ$ using the standard triangle notation.
Hence $\angle AXH = \angle MXH + \angle AXM = \left(\beta + \frac{\alpha}{2}\right) + \frac{\alpha}{2} = \beta + \alpha$ since $\angle AXM = \angle DAC = \frac{\alpha}{2}$ by hypothesis. Hence $\angle AXH + \angle ACH = \beta + \alpha + \gamma = 180^\circ$ and therefore $AXHC$ is cyclic and from this we deduce $\angle AXC = \angle AHC = 90^\circ$ which is what we wanted.

Solution:

Let $A'$ be the intersection, different from $A$, of the circumscribed circle of $AMX$ with the segment $AB$. We have
$$\angle AA'M = \angle AXM = \angle DAC = \angle MAA'$$
where the first equality holds because $AA'XM$ is cyclic, the second holds by hypothesis and the third because $AD$ is the bisector. Hence the triangle $AA'M$ is isosceles. Let $M'$ be the midpoint of $AA'$. Since $AA'M$ is isosceles, $MM' \perp AB$. Moreover, by Thales' theorem, $MM' \parallel DA'$ and hence $DA' \perp AB$.
Let $H$ be the foot of the altitude drawn from $A$ to $BC$. From now on we assume that $H$ lies between $B$ and $D$. The other case, namely $D$ between $B$ and $H$, is treated analogously. Since $DA' \perp AB$ we have $\angle DA'A = \angle DHA = 90^\circ$ and hence the quadrilateral $AA'HD$ is cyclic.
We show that $A'BHX$ is also cyclic. Indeed $\angle BXA' = 180^\circ - \angle A'XM = \angle A'AD$, where the last equality holds because $AA'XM$ is cyclic; $\angle BHA' = 90^\circ - \angle A'HA = 90^\circ - \angle A'DA = \angle A'AD$, where the second equality holds because $AA'HD$ is cyclic and the third because the triangle $AA'D$ is right-angled. Hence $A'BHX$ is cyclic since we have shown
$$\angle BXA' = \angle A'AD = \angle BHA'.$$
Finally we show that $MXHD$ is a cyclic quadrilateral. Indeed
$$\angle XMD = 180^\circ - \angle XMA = \angle AA'X = 180^\circ - \angle XA'B = \angle XHB$$
where the second equality holds because $AA'XM$ is cyclic and the fourth because $A'BHX$ is cyclic.
From the cyclicity of $MXHD$ we deduce $\angle MXH = 180^\circ - \angle MDH$. In the triangle $ABD$ we have $\angle MDH = 180^\circ - \beta - \frac{\alpha}{2}$ using the standard triangle notation. Hence $\angle MXH = \beta + \frac{\alpha}{2}$ and $\angle AXH = \angle MXH + \angle AXM = \left(\beta + \frac{\alpha}{2}\right) + \frac{\alpha}{2} = \beta + \alpha$ since $\angle AXM = \angle DAC = \frac{\alpha}{2}$ by hypothesis. Hence $\angle AXH + \angle ACH = \beta + \alpha + \gamma = 180^\circ$ and therefore $AXHC$ is cyclic and from this we deduce $\angle AXC = \angle AHC = 90^\circ$ which is what we wanted.