By the conditions of the task it follows that the trapezoid is isosceles. Let $K$ be the midpoint of $AD$, and let $\angle CAB = \angle MBC = \varphi$. Then
$$\angle MKA = 180^\circ - \angle KAC = 180^\circ - \angle MBA.$$
Therefore the quadrilateral $ABMK$ is inscribed. Then, by the conditions we have that $\triangle ABD$ is isosceles, from where we get $\angle AKB = 90^\circ$.
Now, because of the fact that $ABMK$ is inscribed, we have $\angle AMB = \angle AKB = 90^\circ$ i.e. we get that the triangle $\triangle AMB$ is a right isosceles triangle.
Let $M_1$ be the foot of the altitude from $M$. Then
$$\overline{MM_1} = \overline{AM_1} = \frac{\overline{AB}}{2} = \frac{\overline{AC}}{2},$$
so we get that $\varphi = 30^\circ$.
Now we easily get $\angle ABC = 30^\circ + 45^\circ = 75^\circ$ and $\angle ADC = 105^\circ$.