<s>[INST] Help to solve this geometry problem 
How to find the length of $BG$ and $EC$ using $\alpha$ and $\beta$ ?               
 [/INST] Writing
$$a := |\overline{BC}| \qquad b := |\overline{CA}| \qquad c := |\overline{AB}|$$
$$y := |\overline{CE}| = |\overline{ED}| \qquad z := |\overline{BG}| = |\overline{GF}|$$
we have, in right triangles $\triangle ADG$ and $\triangle AFE$,

$$\frac{b-2y}{c-z} = \cos A = \frac{c-2z}{b-y}$$

Solving for $y$ and $z$ gives
$$\begin{align}
y &= \frac{2 b - c \cos A - b \cos A^2}{(2 - \cos A) (2 + \cos A)} = 
\frac{(b - c \cos A) + b (1-\cos A^2)}{(2 - \cos A) (2 + \cos A)} =
\frac{a \cos C + b \sin A^2}{(2 - \cos A) (2 + \cos A)} \\
z &= \frac{2 c - b \cos A - c \cos A^2}{(2 - \cos A) (2 + \cos A)} = \frac{(c - b \cos A) + c (1-\cos A^2)}{(2 - \cos A) (2 + \cos A)} = \frac{a \cos B + c \sin A^2}{(2 - \cos A) (2 + \cos A)} 
\end{align}$$
By the Law of Sines, we can write $b = a \sin B/\sin A$ and $c = a \sin C/\sin A$, so that
$$y = a\;\frac{\cos C + \sin A \sin B}{(2 - \cos A) (2 + \cos A)} \qquad z = a\;\frac{\cos B + \sin C \sin A}{(2 - \cos A) (2 + \cos A)}$$
With $A+B+C=\pi$, so that $\sin A = \sin(B+C)$ and $\cos A = - \cos(B+C)$, we get an answer that agrees with @Jack's (noting that his expressions are for $|\overline{BF}| = 2y$ and $|\overline{CD}| =2z$):

$$y = a\;\frac{3\cos C - \cos(2B+C)}{2(4 - \cos^2(B+C))} \qquad z = a\;\frac{ 3\cos B - \cos(B+2C)}{2(4-\cos^2(B+C))}$$

</s>