<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] Use the sine law. Let $x=BD$ and $y=AD$ and $z=CD$ which is what we want to find.Then we have in triangle $ABC$,
$$\frac{\sin 120}{x+y}=\frac{\sin A}{a}=\frac{\sin B}{b}$$ and so
$$\sin A=\frac{\sqrt{3}}{2}\frac{a}{x+y}$$
$$\sin B=\frac{\sqrt{3}}{2}\frac{b}{x+y}$$
Next in triangles $ACD$ and $BCD$ you have 
$$\frac{\sin 60}{y}=\frac{\sin A}{z}$$
$$\frac{\sin 60}{x}=\frac{\sin B}{z}$$
So $$z=\frac{2}{\sqrt{3}} y  \ \sin A = \frac{ay}{x+y}$$
$$z=\frac{2}{\sqrt{3}} x \sin B= \frac{bx}{x+y}$$
multiplying the first by $b$, the second by $a$ and adding,
 $$(a+b)z=ab$$
So I think its B)
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