<s>[INST] 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.
$$4 \sin x \cos 2x= \sin(180 - 7x)$$
 [/INST] 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.
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