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.
 A: Let $x:=\gamma, y:=\theta+\beta_2, p:=\overline{AD}$ and $q:=\overline{BM}=\overline{CM}$.
Then, one can write
$$\angle{ABC}=180-x-y\qquad\text{and}\qquad \angle{BAM}=y-\alpha_2$$
Applying the law of sines to $\triangle{ABC}$, one has
$$\overline{AC}=\frac{2q\sin(180-x-y)}{\sin(y-\alpha_2+\alpha_2)}\tag1$$
Applying the law of sines to $\triangle{AMC}$, one has
$$\overline{AM}=\frac{q\sin x}{\sin\alpha_2}\tag2$$
Applying the law of sines to $\triangle{AMB}$, one has
$$\overline{AM}=\frac{q\sin(180-x-y)}{\sin(y-\alpha_2)}\tag3$$
Applying the law of sines to $\triangle{DBC}$, one has
$$p+\overline{AC}=\frac{2q\sin(\beta_2+180-x-y)}{\sin\theta}\tag4$$
It follows from $(1)(4)$ that
$$p+\frac{2q\sin(180-x-y)}{\sin(y-\alpha_2+\alpha_2)}=\frac{2q\sin(\beta_2+180-x-y)}{\sin\theta}$$
i.e.
$$p+\frac{2q\sin(x+y)}{\sin y}=\frac{2q\sin(x+\theta)}{\sin\theta}$$
Multiplying the both sides by $\sin y\sin\theta$ gives
$$p\sin y\sin\theta+2q\sin(x+y)\sin\theta=2q\sin(x+\theta)\sin y$$
i.e.
$$p\sin y\sin\theta+2q\sin\theta\sin x\cos y+2q\sin\theta\cos x\sin y=2q\sin y\sin x\cos\theta+2q\sin y\cos x\sin\theta$$
i.e.
$$(p\sin\theta-2q\sin x\cos\theta)\sin y+2q\sin\theta\sin x\cos y=0\tag5$$
It follows from $(2)(3)$ that
$$\frac{q\sin x}{\sin\alpha_2}=\frac{q\sin(180-x-y)}{\sin(y-\alpha_2)}$$
i.e.
$$\frac{\sin x}{\sin\alpha_2}=\frac{\sin(x+y)}{\sin(y-\alpha_2)}$$
Multiplying the both sides by $\sin\alpha_2\sin(y-\alpha_2)$ gives
$$\sin x\sin(y-\alpha_2)=\sin\alpha_2\sin(x+y)$$
i.e.
$$\sin x\sin y\cos\alpha_2-\sin x\cos y\sin\alpha_2=\sin\alpha_2\sin x\cos y+\sin\alpha_2\cos x\sin y$$
i.e.
$$(\sin\alpha_2\cos x-\sin x\cos\alpha_2)\sin y+2\sin\alpha_2\sin x\cos y=0\tag6$$
Now, $\sin\alpha_2\times (5)-q\sin\theta\times (6)$ gives
$$\sin\alpha_2(p\sin\theta-2q\sin x\cos\theta)\sin y-q\sin\theta(\sin\alpha_2\cos x-\sin x\cos\alpha_2)\sin y=0$$
Dividing the both sides by $\sin y$ gives
$$S\cos x+T\sin x=U$$
where
$$\begin{cases}S=q\sin\theta\sin\alpha_2\gt 0
\\T=2q\sin\alpha_2\cos\theta-q\sin\theta\cos\alpha_2
\\ U=p\sin\alpha_2\sin\theta\gt 0\end{cases}$$
So, we can write
$$\sqrt{S^2+T^2}\cos(x-V)=U$$
where
$$V=\arcsin\bigg(\frac{T}{\sqrt{S^2+T^2}}\bigg)$$
which satisfies $-90\lt V\lt 90$.
So, one gets
$$x=V\pm\arccos\bigg(\frac{U}{\sqrt{S^2+T^2}}\bigg)$$
where one has to choose $x$ satisfying $0\lt x\lt 180$.
It follows from $(6)$ that
$$y=\begin{cases}90&\text{if $x=\alpha_2$}
\\\\\arctan W_x&\text{if $x\not=\alpha_2$ and $W_x\gt 0$}
\\\\180+\arctan W_x&\text{if $x\not=\alpha_2$ and $W_x\lt 0$}\end{cases}$$
where
$$W_x=\frac{-2\sin\alpha_2\sin x}{\sin(\alpha_2-x)}$$
In conclusion, one gets
$$\color{red}{\begin{align}\gamma&=V\pm\arccos\bigg(\frac{U}{\sqrt{S^2+T^2}}\bigg)
\\\\\beta_2&=\begin{cases}90-\theta&\text{if $\gamma=\alpha_2$}
\\\\-\theta+\arctan W_{\gamma}&\text{if $\gamma\not=\alpha_2$ and $W_{\gamma}\gt 0$}\\\\-\theta+180+\arctan W_{\gamma}&\text{if $\gamma\not=\alpha_2$ and $W_{\gamma}\lt 0$}\end{cases}\end{align}}$$
where
$$\begin{cases}0\lt\gamma\lt 180
\\\max(0,\alpha_2-\theta)\lt \beta_2\lt 180-\theta-\gamma
\\S=\overline{BM}\sin\theta\sin\alpha_2
\\T=2\overline{BM}\sin\alpha_2\cos\theta-\overline{BM}\sin\theta\cos\alpha_2
\\ U=\overline{AD}\sin\alpha_2\sin\theta
\\\displaystyle V=\arcsin\bigg(\frac{T}{\sqrt{S^2+T^2}}\bigg)
\\\displaystyle W_{\gamma}=\frac{-2\sin\alpha_2\sin\gamma}{\sin(\alpha_2-\gamma)}\end{cases}$$
A: By the law of sines in $\triangle ABM$ and $\triangle ABD$, we have
\begin{multline*}
BM \sin (α_2 + γ) = AB \sin α_1 = \frac{AD\sin θ \sin α_1}{\sin β_2} = \frac{AD \sin θ \sin (β_2 + θ - α_2)}{\sin β_2} \\
= AD \sin θ \cos (θ - α₂) + AD \sin θ \sin (θ - α₂) \cot β_2.
\end{multline*}
Meanwhile, by the law of sines in $\triangle ABM$ and $\triangle ACM$, we have
\begin{multline*}
0 = \frac{\sin α_1 \sin γ - \sin α_2 \sin β_1}{\sin β_2} = \frac{\sin (β_2 + θ - α_2) \sin γ - \sin α_2 \sin (β_2 + θ + γ)}{\sin β_2} \\
= \cos (θ - α_2) \sin γ - \sin α_2 \cos (θ + γ) + [\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)] \cot β_2.
\end{multline*}
Subtracting $AD \sin θ \sin (θ - α_2)$ times the second equation from $[\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)]$ times the first equation eliminates $β_2$:
\begin{multline*}
[\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)] BM \sin (α_2 + γ) \\ = [\sin (θ - α_2) \sin γ - \sin α_2 \sin (θ + γ)] AD \sin θ \cos (θ - α_2)
\\ - [\cos (θ - α_2) \sin γ - \sin α_2 \cos (θ + γ)] AD \sin θ \sin (θ - α_2).
\end{multline*}
If we let $t = \tan \frac γ2$, so that $\cos γ = \frac{1 - t^2}{1 + t^2}$, $\sin γ = \frac{2t}{1 + t^2}$, we can expand the above into a quartic equation in $t$, and solve it using the quartic formula. We can then solve for $γ$ and use one of the first two equations to solve for $β_2$.
A: 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 $\beta_2=\pi/4+\pi/4-\pi/3$ $=\pi/6$.
A: 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. $\square$
