Find $x$ in the triangle 

  the triangle without point F is drawn on scale, while I made the point F is explained below

So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, \tan$ answer)
$$\sin \angle ACB = \frac {DF}{CD} = \frac{AB}{BC}$$
$$\cos \angle ACB = \frac {CF}{CD} = \frac{AC}{BC}$$
$$\tan \angle ACB = \frac {DF}{CF} = \frac{AB}{AC}$$
putting known data into it
\begin{align}
\frac {DF}{CD} &= \frac {AB}{12} \quad(1) \\
\frac {EF+3}{CD} &= \frac {2CD}{12} \\
\frac {EF+3}{CD} &= \frac {CD}{6} \quad(2)\\
\frac {DF}{EF+3} &= \frac {AB}{2CD} \quad(3)
\end{align}
I've stuck at here, how do I find their length?

Let $AD=y$ so that $\cos C=\frac{2y}{12}$
Then using the cosine rule, $$x^2=y^2+3^2-6y\cos C$$
So $x=3$
