# Angle trisector segments on side opposite of a triangle

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 $\angle C/n?$

• have you tried anything yourself? – David Quinn Apr 8 '18 at 17:30
• Hoping that more powerful vector approaches might be in existence I avoided a laborious ( inelegant ?) conventional trig/algebraic calculation. – Narasimham Apr 8 '18 at 17:42
• Do you consider it too laborious to coordinatize, with $C = (0,0)$, $A = (b,0)$, $B=(a\cos 3\theta, a\sin 3\theta)$, and then to intersect $\overleftrightarrow{AB}$ with lines $y = x\,\tan k \theta$? – Blue Apr 8 '18 at 17:49
• Not reeally. can I ask for generalization,? Angle $C$ poly-sected ( equal $\angle C/n$ with segments $( c_1,c_2,... c_n )$ in a while. – Narasimham Apr 8 '18 at 19:26
• I was just wondering for $n$ whether segment lengths make a nice set with trig/algebraic quantities. – Narasimham Apr 8 '18 at 19:35

## 1 Answer

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 $c_i$

• Can I now ask to generalize when angle $C$ is divided into $n$ equal parts? – Narasimham Apr 8 '18 at 19:29
• @Narasimham I have added a bit more... – David Quinn Apr 8 '18 at 19:42