Find the bisector in the given condition I need to find x which is bisectrix for that 60°. Given red segments 9 and 12 cm and a circle. It`s seems like school exercise, but i`m I am get stuck, please help (have tried several approaches). Please, give any advices or direction to solve it. Thanks. 
↓ See the image below ↓
Image of the problem


Let 
$|AB|=c=9$,
$|AC|=b=12$,
$|AD|=d=?$,
$\angle CAB=\alpha=60^\circ$,
$\angle DAB=\angle CAD=\tfrac12\,\alpha=30^\circ$.
Then by the cosine rule we have
\begin{align}
a^2&=b^2+c^2-2bc\cos\alpha=b^2+c^2-bc=117
,\\
a&=3\sqrt{13}
.
\end{align} 
The area $S$ and the radius of the circle $R$
of $\triangle ABC$ are
\begin{align} 
S&=\tfrac12bc\sin\alpha=27\sqrt3
,\\
R=|OA|=|OB|=|OC|&=\frac{abc}{4S}
=\sqrt{39}
.
\end{align} 
\begin{align} 
\triangle ABD:\quad
d^2&=c^2+R^2-2cR\,\cos(\beta+30^\circ)
=
120-27\sqrt{13}\,\cos\beta+9\,\sqrt{39}\,\sin\beta
,
\end{align}
\begin{align} 
\triangle ABC:\quad
\cos\beta&=\frac{a^2+c^2-b^2}{2ac}
=\tfrac{\sqrt{13}}{13}
,\\
\sin\beta&=\sqrt{1-\cos^2\beta}=\tfrac{2\sqrt{39}}{13}
,\\
d^2&=147
,\\
d&=7\sqrt3
.
\end{align}
