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?
 A: 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 $x=3$
A: It's 3!  To see it, draw the perpendicular From E to AC (let G be the foot of that perpendicular).  Then triangle EGC is similar to ABC with scale factor $\frac 14$, whence GC is $\frac 12$ of DC.  Hence the two right triangles EGC and EGD are congruent and the result follows.
Note: this was corrected to reflect a typo-generated arithmetic error pointed out by the OP.
A: 
$K$ will be like that so - $DK\perp CD$
we know that $DK\perp CD$ than $DK||AB$ (because $AC\perp AB$), Also we know that $CD=AD$, because of that we can understand $DK$ is median of triangle $\Delta ABC$ so $CK=KB=6$.
We know that $CE=3$, than $CE=EK=3$. $\Delta CDK$ is a right triangle, than we can understand that - $DE=CE=EK=3$ (Because $DE$ is median)
A: Let, $\angle ACB=\alpha$ then we have  $$AC=BC\cos \alpha=(9+3)\cos\alpha=12\cos\alpha$$ $$\implies DC=AD=\frac{AC}{2}=6\cos\alpha$$ 
in right $\triangle DFC$ we have 
$$DF=(DC)\sin\alpha=6\sin\alpha\cos\alpha$$ $$FC=(DC)\cos\alpha=6\cos\alpha\cos\alpha=6\cos^2\alpha$$ $$FE=FC-CE=6\cos^2\alpha-3$$ Now, applying Pythagorean theorem in right $\triangle DFE$ as follows $$(DE)^2=(DF)^2+(FE)^2$$  $$\implies x^2=(6\sin\alpha\cos\alpha)^2+(6\cos^2\alpha-3)^2$$  $$\implies x^2=36\sin^2 \alpha\cos^2\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$ $$=36(1-\cos^2\alpha)\cos^2\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$  $$=36\cos^2\alpha-36\cos^4\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$  $$\implies x^2=9$$ $$\implies \color{blue}{x=3}$$
