What's the length of EC?

Question

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?

Answer 1

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

$$ |E-C| = \frac{28 \sqrt{1240129-291740 \sqrt{6}}}{4113} $$

Answer 2

COMMENT:

It can be seen that $\angle ACB=45^o$:

$AB^2=AC^2+BC^2-2 AC\times BC \times Cos(\angle ACB)$

Putting given values in this relation we get $\angle ACB=45^o$.Now we draw the diameter of a parallelogram that can be constructed on sides AC and BC and mark the intersection of diameter and side AB as E'. We can calculate E'=51.1 as described bellow. As can be seen EC

$(2 E'C)^2=AC^2+BC^2+2 AC\times BC \times Cos(\angle ACB)$

Plugging values we get :

$2 E'C=102.2$ ⇒ $E'C=51.1$