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.

Changing-up notation slightly, I'll use $\alpha$, $\beta$, $\delta$ for OP's $\alpha_2$, $\beta_2$, $\theta$. Also, let $N$ and $P$ be the feet of the perpendiculars from $A$ to $\overline{BD}$ and $M$ to $\overline{AC}$, and let $\square DPQR$ be a rectangle, with $B$ on $\overline{QR}$. Finally, define
$$a := |BM|=|MC| \qquad d := |AD| \qquad p := |MP|=|QM|$$
I'll also rotate the figure to make $\overline{DC}$ horizontal:

For the configuration shown, straightforward right-triangle trigonometry gives the (signed) lengths of various segments. Equating opposite sides of $\square DPQR$, and invoking the Pythagorean Theorem in $\triangle CMP$, gives these three equations:
$$\begin{align}
2p &= d(\cot\beta+\cot\delta) \sin^2\delta \quad\left(= d\frac{\cot\beta+\cot\delta}{1+\cot^2\delta}\right) \tag1 \\
d+p \cot\alpha &= 2p\cot\delta+p\cot\gamma \tag2 \\[0.5em]
a^2 &= p^2 + p^2 \cot^2\gamma \tag3
\end{align}$$
Equations $(1)$ and $(2)$ give expressions involving our target angles. Defining
$$u := \cot\alpha \qquad v := \cot\delta \qquad \lambda := u-2v$$
we have
$$
\cot\beta = \frac{2p(1+v^2)-dv}{d} \qquad\qquad
\cot\gamma = \frac{d+p\lambda}{p}\tag4$$
Substituting $\cot\gamma$ into $(3)$ gives a quadratic in $p$ that we solve as
$$p = \frac{-d\lambda\pm\sqrt{a^2(1+\lambda^2)-d^2}}{1+\lambda^2} \tag5$$
discarding non-positive root(s). Substituting from $(5)$ back into $(4)$ determines $\beta$ and $\gamma$. $\square$
(Considering whether there can be two positive roots $p$ —hence, two overall solutions— is left as an exercise to the reader.)

As something of a sanity check ... Consider the isosceles right triangle example in @Anatoly's answer.

We have
$$\alpha=45^\circ \qquad \delta=60^\circ \qquad a = 1 \qquad |AB|=\sqrt{2} \qquad d = \frac{\sqrt{6}}{3} $$
so that
$$u = 1 \qquad v = \frac{\sqrt{3}}3 \qquad \lambda = \frac13(3-2\sqrt{3}) \qquad\to\qquad p =\frac{\sqrt{2}}{2}$$
(Note: Reducing $(5)$ to $\sqrt{2}/2$ is a non-trivial exercise!) Then,
$$\cot\beta = \sqrt{3} \qquad \cot\gamma = 1 \qquad\to\qquad \beta=30^\circ \qquad \gamma=45^\circ$$
as expected. $\square$
