9. Consider a triangle $A B C$ and a point $D$ on the side $A C$. If $\overline{A B}=\overline{D C}=1, \angle D B C=30^{\circ}$ and $\angle A B D=90^{\circ}$, calculate the value of $\overline{A D}$.
Solution. Let $\angle A D B=\alpha$, then we have $\angle B D C=180-\alpha$. Using the Law of Sines in triangle $A D B$ we have that

$$
\frac{x}{1}=\frac{1}{\sin \alpha}=\frac{B D}{\sin (90-\alpha)}
$$

and in triangle $D B C$ we will have that

$$
\frac{1}{\sin 30}=\frac{B D}{\sin (\alpha-30)}=\frac{B C}{\sin (180-\alpha)}
$$

From the first equalities we deduce that $B D=\frac{\cos \alpha}{\sin \alpha}$ and from the second that

$$
B D=\frac{\sin (\alpha-30)}{\sin 30}=2 \sin (\alpha-30)=2(\sin \alpha \cos 30-\cos \alpha \sin 30)
$$

and thus, $B D=\sqrt{3} \sin \alpha-\cos \alpha$. Equating we have

$$
\frac{\cos \alpha}{\sin \alpha}=\sqrt{3} \sin \alpha-\cos \alpha \Rightarrow \cos \alpha=\sqrt{3} \sin ^{2} \alpha-\sin \alpha \cos \alpha
$$

from which $\cos \alpha(1+\sin \alpha)=\sqrt{3} \sin ^{2} \alpha$ and, squaring and denoting by $t=\sin \alpha,\left(1-t^{2}\right)(1+t)^{2}=3 t^{4}$, from which $4 t^{4}+2 t^{3}-2 t-1=0$ and, factoring, $2 t^{3}(2 t+1)-(2 t+1)=0$. Therefore, $\left(2 t^{3}-1\right)(2 t+1)=0$ and, therefore, we have two options: If $2 t+1=0$, then $t=-\frac{1}{2}$ and $x=1 / t=-2$, impossible. Therefore, $2 t^{3}-1=0$, so $t=\frac{1^{2}}{\sqrt[3]{2}}$ and we deduce that $x=\sqrt[3]{2}$.