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

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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}