Calculate the value of angle $ACB$ In an ABC triangle the angle $BAC$ is twice the angle $ACB.$ Consider a point $D$ in segment $AC$ so that the angle $DBA$ is twice the angle $BAD.$ Calculate the value of angle $ACB,$ knowing that the measurement of segment $CD$ is equal to the sum between twice the measurement of segment $BD$ and the length of segment $AD.$
Attemp:After using the law of sines on triangles ABD and BCD, I got the weird-looking equation attached.
I think my approach most likely is not correct.
$$4 \sin x \cos 2x= \sin(180 - 7x)$$
 A: Denoting $\angle ACB=\gamma$ (the one we need to find), you have that
$$CD=2BD+AD\Leftrightarrow \\
\frac{CD}{BD}=2+\frac{AD}{BD}\quad (1) $$
Apply sine law in $\triangle ADB$ and $\triangle CDB$:
$$\frac{CD}{BD}=\frac{\sin7\gamma}{\sin\gamma}\\ \frac{AD}{BD}=\frac{\sin4\gamma}{\sin2\gamma}=2\cos2\gamma  $$
so let $\sin\gamma=x$ and substitute in $(1)$. $\cos2\gamma=1-2x^2$ and $\sin7\gamma=7x-56x^3+112x^5-64x^7$ (see here).
You get the equation
$$64x^6-112x^4+52x^2-3=0\Leftrightarrow \\
(4x^2-3)(16x^4-16x^2+1)=0 $$
which is solvable by letting $t=x^2$. And you have to take into account that $7\gamma<180^{\circ}$, so $0<x<\sin\frac{180^{\circ}}{7}<\sin\frac{180^{\circ}}{6}=\frac 12\Rightarrow 0<t<\frac 14$. We get
$t=\frac{2-\sqrt{3}}{4}\Rightarrow x=\frac{\sqrt{2-\sqrt{3}}}{2} \Rightarrow \gamma=15^{\circ}$ (see here for a table of trig. values).
A: An euclidean-trigonometric cocktail
Take on $DC$ a point $E$ such that $ED\cong BD$. By our hypotheses (that can be rewritten as $CD - BD \cong AD + BD$) we have that
$$CE\cong AE.$$
Produce $AB$ to $F$ so that $\triangle AEF$ is isososceles.
Call now $\angle CAB = x$, for simplicity. Then of course $\angle ABD = 2x$ and $\angle ACB = \frac{x}2$. 


*

*$A$, $C$, and $F$ lie on the circle centered in $E$ with radius $\frac{\overline{AC}}2$, therefore $AF\perp CF$.

*Angle chasing yields $\angle BEF \cong\angle EBF=\frac{\pi}2-\frac{x}2$, therefore $BF\cong \frac{AC}2$.

*External angle theorem yields $\angle CBF = \frac{3x}2$.



We must have 
$$\overline{AC} \sin x \cot \frac{3x}2 = \frac{\overline{AC}}2.$$
Trigonometric manipulation gives:
\begin{eqnarray}
\frac{\cos\frac{3x}2\sin x}{\sin\frac{3x}2} &=& \frac12\\
2\frac{\cos \frac{x}2\cos x - \sin\frac{x}2\sin x}{\sin \frac{x}2\cos x + \cos\frac{x}2\sin x}\sin\frac{x}2\cos\frac{x}2 &=& \frac12\\
2\frac{\cos\frac{x}2\left(2\cos^2\frac{x}2-1\right)-2\sin^2\frac{x}2\cos\frac{x}2}{\sin\frac{x}2\left(2\cos^2\frac{x}2-1\right)+2\cos^2\frac{x}2\sin\frac{x}2}\sin\frac{x}2\cos\frac{x}2 &=& \frac12\\
\frac{2\cos^2\frac{x}2\left(2\cos x-1\right)}{2\cos x +1}&=&\frac12\\
\frac{(\cos x+1)(2\cos x-1)}{2\cos x + 1}&=&\frac12,
\end{eqnarray}
which in turns yields
$$4\cos^2x - 3 = 0,$$
and thus $x = \frac{\pi}6$ as the only geometrically valid solution to the problem.
A: A purely euclidean path
Here is an approach based solely on congruences. It shares part of the path shown in my previous answer. I will repeat all the steps, anyway, to make this answer self-standing. Let $\angle CAB = \alpha$, so that $\angle ABD = 2\alpha$ and $\angle ACB = \frac{\alpha}2$.



*

*Draw $E$ on $CD$ so that $BD\cong DE$; the hypothesis $CD \cong 2BD + AD$ implies $$AE\cong EC.$$

*Produce $AB$ to $F$ so that $$AE\cong EF.$$Since $A$, $C$, and $F$ lie on the half-circle centered in $E$ and with radius $\frac{\overline{AC}}2$, we have $CF\perp AF$. Produce $CF$ to $L$, so that $CF\cong FL$. 

*Taking advantage of the fact that $\triangle BDE$ and $\triangle ECF$ are isosceles we obtain that $\angle EBF \cong \angle BEF = \frac{\pi}2-\frac{\alpha}2$. So $$BF \cong EF.$$

*$\angle BCF = \frac{\pi}2-\frac{3\alpha}2$, and $\angle BEC = \frac{\pi}2+\frac{3\alpha}2$ by angle chasing.

*$\triangle BFC \cong \triangle BFL$ by SAS criterion, implying in particular that $\angle BLF \cong \angle BCF = \frac{\pi}2-\frac{3\alpha}2.$

*Points 4. and 5. imply that $\square CEBL$ is cyclic, because $\angle BEC$ and $\angle BLC$ are supplementary. Since $AF \perp CL$, and $F$ is the midpoint of $CL$, the center of its circumscribed circle must lie on $AF$. By 3., the center is $F$. So $CF\cong \frac{AC}2$ and the thesis, i.e. $$\boxed{\alpha = \frac{\pi}6},$$ follows immediately.

