<s>[INST] What's the length of EC? Consider the following construction (the lenght of segment EC in the image is just an approximation):

I wanted to calculate the exact value of the length of EC.
Using the law of cosines, law of sines, the angle bisector theorem and some manipulation I arrived at: 
$$\sqrt{16+\left(5-\frac{35 \csc \left(\frac{\pi }{4}+\cos ^{-1}\left(\frac{29}{35}\right)\right)}{9 \sqrt{2}}\right)^2+\frac{8}{5} \left(5-\frac{35 \csc \left(\frac{\pi }{4}+\cos ^{-1}\left(\frac{29}{35}\right)\right)}{9 \sqrt{2}}\right)}$$ which simplifies to $\displaystyle \frac{28 \sqrt{1240129-291740 \sqrt{6}}}{4113}$.
While I believe this is correct I also believe there should be a easier way to solve this. Is there a simpler argument? 
 [/INST] Calling $A = \{x_a,y_a\}, B = \{0,0\}, C = \{x_c,0\}$ and equating
$$
\cases{
|A-B|^2 = 5^2\\
|B-C|^2 = 7^2\\
|C-A|^2 = 4^2
}
$$
we obtain
$$
\cases{
A = \{\frac{29}{7},-\frac{8\sqrt 6}{7}\}\\
C = \{7,0\}
}
$$
Calling now
$$
\vec v_{AC} = \frac{C-A}{|C-A|} =\left \{\frac{5}{7},\frac{2\sqrt 6}{7}\right\}\\
\vec v_{AB} = \frac{B-A}{|B-A|} =\left \{-\frac{29}{35},\frac{8\sqrt 6}{35}\right\}
$$
we have for the point $D$ determination
$$
A + \lambda (\vec v_{AC}+\vec v_{AB}) = B + \mu(C-B)
$$
which solved for $\lambda,\mu$ gives
$$
D = \left\{\frac{35}{9},0\right\}
$$
and finally the point $E$ determination
$$
A + \lambda(B-A) = D + \mu\{-1,-1\}
$$
which solved for $\lambda,\mu$ gives
$$
E = \left\{\frac{1015 \left(29-8 \sqrt{6}\right)}{4113},-\frac{560}{144+87 \sqrt{6}}\right\}
$$
hence
$$
|E-C| = \frac{28 \sqrt{1240129-291740 \sqrt{6}}}{4113}
$$
</s>