<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] 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 $$(DE)^2=(DF)^2+(FE)^2$$  $$\implies x^2=(6\sin\alpha\cos\alpha)^2+(6\cos^2\alpha-3)^2$$  $$\implies x^2=36\sin^2 \alpha\cos^2\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$ $$=36(1-\cos^2\alpha)\cos^2\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$  $$=36\cos^2\alpha-36\cos^4\alpha+36\cos^4\alpha+9-36\cos^2\alpha$$  $$\implies x^2=9$$ $$\implies \color{blue}{x=3}$$
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