Angle Bisector Theorem (2) 
We have a $\triangle ABC$ with $\angle ACB=120^\circ$. $CL$ and $AT$ are angle bisectors, $L\in AB$ and $T\in BC$. I should show that $LT$ is the angle bisector of $\angle CLB$.
  

By the angle bisector theorem, $\dfrac{AL}{LB}=\dfrac{AC}{BC}$ and $\dfrac{CT}{TB}=\dfrac{AC}{AB}$. We should show $\dfrac{CT}{TB}=\dfrac{CL}{LB}$. How can I use $\angle ACL=\angle BCL=60^\circ$? Thank you in advance!
 A: Observe a triangle $ACL$ and extend halfline $AC$ to $X$. 
Notice that $CT$ is outer angle bisector for angle $\angle ACL$ since $\angle LCT = \angle LCX = 60^{\circ}$. 
Since $T$ is also on angle bisector $AT$ of angle $\angle CAL$. So $T$ is at equal distance to lines $CL$ and $LB$, so $LT$ is angle bisector of angle $\angle CLB$.
A: By the Law of sines,
$$\frac{|AB|}{\sin(120^{\circ})}=\frac{|BC|}{\sin(A)},\quad
\frac{|AL|}{\sin(60^{\circ})}=\frac{|CL|}{\sin(A)}.$$
Since $\sin(120^{\circ})=\sin(60^{\circ})$, it follows that
$$\frac{|AB|}{|BC|}=\frac{|AL|}{|CL|}.$$
Hence, from your work,
$$\frac{|CT|}{|TB|}=\frac{|AC|}{|AB|}=\frac{|AC|}{|BC|}\cdot
\frac{|CL|}{|AL|}=\frac{|AL|}{|LB|}\cdot
\frac{|CL|}{|AL|}=\frac{|CL|}{|LB|}$$
and we are done.
P.S. In order to avoid the Law of sines, note that 
$$|AC||BC|\sin(120^{\circ})=2\text{Area}(ABC)=|AB||AC|\sin(A)$$
implies 
$$\frac{|AB|}{\sin(120^{\circ})}=\frac{|BC|}{\sin(A)}.$$
Similarly, by considering the triangle $ACL$, we find
$$\frac{|AL|}{\sin(60^{\circ})}=\frac{|CL|}{\sin(A)}.$$
