Solve a triangle given one angle of the median, and a side and an angle of its external triangle This is a modified version of this question, and as such I'm using similar wording and visuals. Given:

*

*$α_2$, one of the two angles which the vertex $A$ is split by the median $m$;

*$\overline{AD}$, the length of the segment of a triangle external to $ABC$;

*$\overline{BM}$, the length of half of the side split by the median;

*$θ$, the angle opposite to the shared side $\overline{AB}$;

Find the value of the angles $γ$ and $b_2$.

Similarly to the original question, the proportion $\frac{\sin{α_1}}{\sin{α_2}}=\frac{\sin{β_1}}{\sin{γ}}$ holds true. And it's still true that $β_1 = 180 - α_1 - α_2 - γ$. However, unlike the original scenario, we don't know the value of $α_1$.
With the external angle theorem, we know that $α_1 = θ + β_2 - α_2$, but now we have the $β_2$ variable to resolve.
I figured we could use the law of sines to establish $\frac{\sin{β_2}}{\overline{AD}}=\frac{\sin{θ}}{\overline{AB}}$, but after a few variable swaps the problem ultimately seems to loop back to $α_1$.
I feel that this problem is solvable, as the given parameters uniquely identify the two triangles. But I can't find out how to take it further than I have.

To better highlight the unknown terms, let us set $\alpha_1=x$, $\beta_1=y$, $\gamma=z$. The known quantities are $\alpha_2$, $AD$, $BM$, $\theta$.
The first two equations, already reported in the OP,  are:
$$
\frac{\sin{x}}{\sin{α_2}}=\frac{\sin{y}}{\sin{z}}
\tag{1}$$
$$y = 180 - x - α_2 - z \tag{2}$$
As we have $3$ unknowns, we need a third equation.

From the triangle $\triangle{ABD}$ we have:
$$\frac{AB}{\sin \theta}=\frac{AD}{\sin \beta_2}$$
Here we can try to express $AB$ and $\beta_2$ as functions of $x,y,z$ and the other known terms. Considering the triangle  $\triangle{ABC}$, we have
$$AB=\frac{2BM}{\sin(x+\alpha_2)} \,\sin z$$
Also, by the exterior angle theorem, $$\beta_2=x+\alpha_2-\theta$$
Substituting, we obtain our third equation, which complete the system:
$$\frac{2\,BM \, \sin z   }{\sin(x+\alpha_2)} \,=  \frac{AD \, \sin \theta  }{\sin (x+\alpha_2-\theta)} \tag{3}$$
The system can then be solved by the usual methods. Once we obtained $x,y,z$, we can also calculate $\beta_2$ using the formula above.

To provide an example of how the system works: let us consider the very simple case where the triangle $\triangle{ABC}$ is right and isosceles with hypotenuse equal to $2$, and the triangle $\triangle{ABD}$ is right with $\theta=60°$. In this case, the known terms are $\alpha _2=\pi/4$, $\gamma=\pi/3$, $BM=1$, and $AD=1/\sqrt{6}$. By construction, this case corresponds to the trivial case where $x=y=z=\pi/4$ and $\beta_2=\pi/6$.
As expected, the system provides the correct $x,y,z$ solutions, as shown here. From this, we also easily get $\beta_2=\pi/4+\pi/4-\pi/3$ $=\pi/6$.
