Find angle $\alpha$ in $\triangle ABC$ As title suggests, the goal is to find $\alpha$ from the given $\triangle ABC$ with some given angles and sides. I'm posting this here to look for better solutions, as my own, which I'll post as well, is rather complicated.

 A: I think my solution is quite simple. Since $\triangle BDC$ is isosceles, it follows that $\angle BCD = \angle CBD = 2 \alpha$, and as such, $\angle BCA = 3\alpha$. We then compute that
$$
\angle ABD = 180 - \angle CBD - \angle BCA - \angle BAC = 180 - 2\alpha - 3\alpha - 7\alpha = 180 - 12 \alpha.
$$
Because also $\triangle ABD$ is isosceles, we decude from this that
$$
\angle BAD = \frac{180 - \angle ABD}{2} = 6\alpha.
$$
In particular, we find that $\angle CAD = 7\alpha - 6\alpha = \alpha$. This implies that also $\triangle CDA$ is isosceles; whence $|AD| = |DC| = |DB| = |BA|$. It follows that $\triangle BDA$ is even equilateral and as such, its angles are all $60^{\circ}$. Hence $6\alpha = 60^{\circ}$, and thus $\alpha = 10^{\circ}$.
A: So, this'll be my approach to this problem. I'll add an explanation of the process as well as my motivation for certain decisions as best as I can:

1.) Label the $\triangle ABC$ and mark all the appropriate angles and points, it is already given that $AB=BD=DC$. Locate point $E$ outside $\triangle ABC$ and connect it with point
$C$ such that $CE=DC=BD=AB$. It is obvious that $\triangle DCE$ is isosceles, therefore $\angle CDE=\angle CED=\alpha$.
2.) Connect points $A$ and $E$ via segment $AE$ to form
$\triangle AEB$. Locate another point $F$ outside $\triangle AEB$ and connect it to $E$ via $EF$ such that $\angle CEF=2\alpha$ and segment $EF=CB$, next join $C$ and $F$ via $CF$. Notice that $\triangle ECF$ is congruent to $\triangle CDB$ via the SAS property. Therefore, $CF=CE=DC=BD=AB$ and $\angle CFE=2\alpha$. Connect point $F$ with $B$ via
$BF$ to form $\triangle BFC$. Notice that
$\angle FCB=4\alpha$ via the external angle theorem. Further, connect points $A$ and $F$ via segment $AF$.
3.) Notice that $\angle ACF=\angle CAB=7\alpha$. Also notice that, as established before, $CF=AB$. This implies that
$\triangle ABC$ and $\triangle FCA$ are also congruent via the SAS property. This proves that segment $BC=AF=FE$. This also implies that $\triangle AFE$ is isosceles. Now, notice that quadrilateral $ABFC$ has two equal diagonals $AF$ and $BC$, and two equal non-parallel sides $CF$ and $AB$. This proves that this quadrilateral is an isosceles trapezoid, which means that it is in fact a cyclic quadrilateral.
4.) Above implies that $\angle FAB=\angle BCF=4\alpha$. This implies that $\angle FAC=\angle BAC-\angle BAF$. Therefore,
$\angle FAC=7\alpha-4\alpha=3\alpha$. Notice that this means that, in the isosceles triangle $\triangle AFE$,
$\angle FAC=\angle FCD=3\alpha$. This proves that $\angle DEA=\angle CAE=\alpha$. This also proves that segment $AC=DE$. Notice that this implies $\triangle DEF$ is congruent to $\triangle FAC$ and $\triangle ABC$ via the SAS property. This also implies that segment $DF=CF=CE=DC=BD=AB$. But notice that this means $\triangle DCF$ is equilateral with
$\angle DCF=6\alpha$. This means that $6\alpha=60$, therefore $\alpha=10$.
A: I propose a simpler solution.

First step: Let us establish that
$$DA=DC\tag{1}$$
With notations of the figure, in isosceles triangle $ABD$:
$$\gamma = 180°-2(7\alpha - \beta)$$
Let us write now that the sum of angles in triangle $ABC$ is $180°$:
$$7 \alpha + [180°-2(7\alpha - \beta)+2 \alpha]+[2\alpha+\alpha]=180°$$
As a consequence $\beta = \alpha$ ; therefore triangle $ADC$ is isosceles ; as a consequence, we can conclude to (1).

Second step: Due to (1), $D$ is the center of the circumscribed circle to triangle $ABC$. The central angle $\widehat{BDC}$ being the double of the inscribed angle $\widehat{BAC}$, (see here), we have:
$$\widehat{BDC}=14 \alpha $$
As the sum of angles in triangle $BDC$ is 180°, we can write:
$$2 \alpha + 2 \alpha + 14 \alpha = 180°.$$
Conclusion:
$$\alpha = 10°$$
A: Law of Sine on ΔABC
$\displaystyle \frac{AB}{\sin(2α+α)} = \frac{BC = 2\,BD\,\cos(2α)}{\sin(7α)}$
Given $AB=BD$, it simplified to:
$\sin(7α) = 2\sin(3α)\cos(2α) = \sin(5α) + \sin(α)$
$\sin(α) = \sin(7α) - \sin(5α) = 2\cos(6α)\sin(α)$
$\sin(α)≠0$, otherwise no triangle $\displaystyle \quad → \cos(6α) = \frac{1}{2} \quad → α = \frac{60°}{6} = 10°$
