We know the bisector length of angle $C$ is $\dfrac{ 2ab \cos C/2 }{a+b};$
In triangle ACB the $\angle C$ is trisected making segments $(c_1,c_2,c_3) $ on the side opposite. Please help to express these three segments in terms of $ (a,b,c, A,B,C).$
EDIT1:
Can I now ask also now to find lengths of $(c_1,c_2... c_n)?$ when $\angle C$ is poly-sectioned $\angle C/n?$
If the lines of trisection meet $AB$ in points $P$ and $Q$, then we can apply the sine rule in triangle $CPB$ and obtain $$c_1=\frac{a\sin\frac 13C}{\sin(\frac 13C+B)}$$
For $c_3$ you can exchange $B$ for $A$ and $a$ for $b$.
For $c_2$ you can use $c=c_1+c_2+c_3$
To generalise further, suppose the angle $C$ is divided into $n$ equal parts, creating segments $c_1,c_2,...c_n$ on the side $AB$
Let $u_r=c_1+c_2+...+c_r$
Then by exactly the same reasoning as before, $$u_r=\frac{a\sin\frac rnC}{\sin(\frac rnC+B)}$$
From this, one could deduce the $c_i$