Help for a problem with inscribed triangles If we have a triangle $ABC$ with $AB = 3\sqrt 7$, $AC = 3$, $\angle{ACB} = \pi/3$, $CL$ is the bisector of angle $ACB$, $CL$ lies on line $CD$ and $D$ is a point of the circumcircle of triangle $ABC$, what's the length of $CD$?
Here I attach my solution. The problem is that I get $CD = 5\sqrt 3$ while in my text book the solution is given as $4\sqrt 3$ and I really cannot understand where I'm doing wrong. Could somebody help me out?


As you wrote the radius $R$ of the circumcircle of $\triangle{ABC}$ is $\sqrt{21}$. 
Since the radius of the circumcircle of $\triangle{ADC}$ is also $R=\sqrt{21}$, by the law of sines, 
$$2R=\frac{AD}{\sin{\angle{ACD}}}\Rightarrow AD=\sqrt{21}.$$
Then, applying the law of cosines to $\triangle{ACD}$ gives you 
$$21=3^2+CD^2-2\cdot 3\cdot CD\cdot \cos{\angle{ACD}}\Rightarrow CD=4\sqrt 3.$$
