Extensions of a triangle

$\triangle DAC$ is isosceles with $AD=AC=1$ and $\angle{DAC}=120°\angle{ACD}=\angle{ADC}=30°\Rightarrow \angle DCB=90°$ and $\triangle DCB$ is $30°-90°-60°\therefore DC=\sqrt{3}$ .

Then extend $DC$ and draw a parallel by $E$ to $BC$ which cuts $DC$ , as image shows. $\triangle CEG$ is $30°-60°-90°$ with $CE=1\therefore CG=\frac{\sqrt{3}}{2}$ and $GE=\frac{1}{2}$ .

$\triangle DCF\sim \triangle DGE \Rightarrow \frac{DC}{DG}=\frac{CF}{GE} \rightarrow \frac{\sqrt{3}}{\frac{3\sqrt{3}}{2}}=\frac{CF}{\frac{1}{2}} \Rightarrow CF=\frac{2}{3}\times \frac{1}{2}=\boxed{\frac{1}{3}\approx 0.333}$

Let, $\boldsymbol {\angle CEF=\theta}$ & $\boldsymbol {\angle CFE=\alpha}$

From the law of cosine we get, $\boldsymbol {DE=\sqrt{1^2+2^2+2\times1\times2\cos{120^{\circ}}}=\sqrt7}$ From the law of sine we get, $\boldsymbol{\frac{\sqrt7}{\sin{120^{\circ}}}=\frac{1}{\sin{\theta}}}$ $\boldsymbol {\Rightarrow \theta=19.11^{\circ}}$ So, $\boldsymbol {\alpha=180^{\circ} -(60^{\circ} +19.11^{\circ} ) =100.89^{\circ} }$ Applying the law of sine one last time we get, $\boldsymbol {\frac{CF} {\sin{19.11^{\circ}}}=\frac{1}{\sin{100.89^{\circ}}}}$ $\boldsymbol {\Rightarrow CF=\frac{\sin{19.11^{\circ}}}{\sin{100.89^{\circ}}} \approx\color{#69047E} {\boxed{0.333}} }$

Draw a parallel by BF from A to DE at M. Then you see 2AM=BF & 2BF=AM. so you will see BF = 0.333.

Applying the theorem of Menelaus of Alexandria ( Menelaus' Theorem ), we have

$\dfrac{AE}{EC}\cdot \dfrac{CF}{FB}\cdot \dfrac{BD}{AD}=1$

$\dfrac{2}{1}\cdot \dfrac{CF}{1+CF}\cdot \dfrac{2}{1}=1$

$\dfrac{4CF}{1+CF}=1$

$3CF=1$

$CF=\dfrac{1}{3}\approx \color{#D61F06}\boxed{0.3333333}$

$Join~ DC.~~~In~\Delta~DBC, ~~~DB=2,~~BC=1,~~\angle~DBC=60.~~~\therefore~\angle~BCD=90,~and~DC=\sqrt3.\\ \Delta~DCE,\\ DC=\sqrt3,~~CE=1,~~\angle~DCE=90+60=150. \\ Let~\angle~EDC=\alpha.~so~\angle~ CED=30-\alpha.\\ By~Sin~Rule~\dfrac{Sin(30-\alpha)} {\sqrt3}=\dfrac{Sin(\alpha)} 1.\\ Expanding~Sin~of~sum~and~simplifying,~~~Cot(\alpha)=3\sqrt3=\dfrac{DC}{CF} .\\ \implies~CF=\dfrac{\sqrt3}{3\sqrt3}=\Large \color{#D61F06}{.333333}$