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?


Your problem is in stating that $\cos \beta = \frac {\sqrt {7}} {14}$. This is incorrect.
Although it is true that $\cos^2 \beta = \frac {7} {196}$, there are two possible values for $\cos \beta$. If you were to make a scale drawing, you would find that AB is over twice as long as Ac, making $\beta$ an obtuse angle.
Thus $\cos \beta = -\frac {\sqrt {7}} {14}$
