<s>[INST] Find length of $CD$ where $\measuredangle BCA=120^\circ$ and $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$ $ABC$ is a triangle with $BC=a,CA=b$ and $\measuredangle BCA=120^\circ$. $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$. Then the length of $CD$ is ____ ?
A)$\frac{a+b}{4}$
B)$\frac{ab}{a+b}$
C)$\frac{a^2+b^2}{2(a+b)}$
D)$\frac{a^2+ab+b^2}{3(a+b)}$
I have tried using $AD+DB=AB$ and using cosine rule on each term. However it gets lengthy.
 [/INST] Solution.    
$\triangle ABC=\triangle ACD+\triangle BCD$    
$\Longrightarrow$ $~\dfrac{ab\sin 120^{\circ}}{2}=\dfrac{b\cdot CD\sin 60^{\circ}}{2}+\dfrac{a\cdot CD\sin 60^{\circ}}{2}$    
$\therefore$ $CD=\dfrac{ab}{a+b}$
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