<s>[INST] 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?
 [/INST] 
$K$ 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)
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