Find length of $CD$ where $\measuredangle BCA=120^\circ$ and $CD$ is the bisector of $\measuredangle BCA$ meeting $AB$ at $D$

Question

$ABC$ 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.

Answer 1

Solution.

$\triangle ABC=\triangle ACD+\triangle BCD$

$\Longrightarrow$ $~\dfrac{ab\sin 120^{\circ}}{2}=\dfrac{b\cdot CD\sin 60^{\circ}}{2}+\dfrac{a\cdot CD\sin 60^{\circ}}{2}$

$\therefore$ $CD=\dfrac{ab}{a+b}$

Answer 2

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)