In $\triangle ABC$, $D$ is a point on side $BC$ that $\angle BAD = \angle CAD =\angle ABC$. If $BD=1$ and $DC=2$, what would be the length of $AB$? 
In $\triangle ABC$, $D$ is a point on side $\overline{BC}$ that $\angle BAD = \angle CAD =\angle ABC$. If $\overline{BD}=1$ and $\overline{DC}=2$, what would be the length of $\overline{AB}$ ?


Things I have done: As $\overline{AD}$ is angle bisector this was the first thing came into my mind $$\frac{\overline{BD}}{\overline{DC}}=\frac{1}{2}=\frac{\overline{AB}}{\overline{AC}} \Rightarrow \frac{\overline{AC}}{2}=\overline{AB}$$
Applying Law of sines gives $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{\overline{AB}}{\overline{AC}} \Rightarrow \frac{\sin{ \angle C}}{\sin{\angle B}}\times \overline{AC} = \overline {AB}$$
thus $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{1}{2}$$
taking $\angle B = x$ results $$\frac{\sin{ \angle C}}{\sin{\angle B}}=\frac{\sin{(180-3x)}}{\sin{(x)}}=\frac{\sin{(3x)}}{\sin{(x)}}=\frac{1}{2}$$
And i stuck here. I solved this question previously by adding some elements and using symmetry and ... . but now I want to solve it without adding any element and constructions.Thanks.
Answer According to answer key: $\frac{\sqrt 6}{2}$ 
 A: Let $\alpha=\angle ABC$.
Since $AD$ is the bisector of $\angle CAB$, we have
$$AB:AC=BD:CD=1:2\Rightarrow AC=2AB.$$
So, since 
$$\angle ACB=180^\circ-3\alpha\Rightarrow \sin(\angle ACB)=\sin(3\alpha)=3\sin \alpha-4\sin^3\alpha,$$we have with $\sin\alpha\not=0$, by the law of sines,
$$\begin{align}\frac{AB}{\sin (\angle ACB)}=\frac{AC}{\sin(\angle ABC)}&\Rightarrow \frac{AB}{3\sin\alpha-4\sin^3\alpha}=\frac{2AB}{\sin\alpha}\\&\Rightarrow \sin\alpha(8\sin^2\alpha-5)=0\\&\Rightarrow \cos^2\alpha=1-\sin^2\alpha=1-\frac 58=\frac 38.\end{align}$$
Hence, we have
$$AB=2\cos \alpha=2\sqrt{\frac 38}=\frac{\sqrt 6}{2}.$$
A: You are correct that the Angle Bisector Theorem gives you that $|AC| = 2|AB|$. Let's take that as our jumping-off point, assigning $x$ and $y$ to lengths as shown:

(I'm using "$x$" instead of "$1$", so that we can follow the value through the formulas better. "$1$"s are so easy to lose!)
By the Law of Cosines in $\triangle ABD$:
$$\begin{align}
y^2 &= x^2 + x^2 - 2 \cdot x\cdot x \cos \delta \\
\implies \qquad y^2&= 2 x^2 - 2 x^2 \cos\delta \qquad (\star)
\end{align}$$
and in $\triangle ACD$:
$$\begin{align}
(2y)^2 &= x^2 + (2x)^2 - 2\cdot x \cdot (2x) \cos(180^\circ-\delta) \\
\implies \qquad 4y^2 &= 5 x^2- 4x^2\cos(180^\circ-\delta) \\
\implies \qquad 4 y^2 &= 5 x^2 + 4 x^2\cos\delta \qquad (\star\star)
\end{align}$$
Then $2(\star) + (\star\star)$ eliminates the $\cos\delta$ terms, so that:
$$ 6 y^2 = 9 x^2 \quad\implies\quad 2 y^2 = 3 x^2 \quad\implies\quad y = \sqrt{\frac32}\;x = \frac{\sqrt{6}}{2}\;x$$
(where we ignore "$\pm$" issues, as we presume $x$ and $y$ to be non-negative).

The fact that the final relation can be written as
$$\frac{\sqrt{3}}{2} x = \frac{\sqrt{2}}{2} y \qquad\text{that is,}\qquad x\sin 60^\circ = y \sin 45^\circ$$
makes me think that there's a more-clever solution, but I don't see it at the moment.
