Help to solve this geometry problem 
How to find the length of $BG$ and $EC$ using $\alpha$ and $\beta$ ?               
 A: Trivially, we have that $FDEG$ is a cyclic quadrilateral, hence:
$$ AF \cdot AG = AD \cdot AE. \tag{1}$$
Let $AF=x,AD=y,AB=c,AC=b$. 
Since $G$ is the midpoint of $BF$ and $E$ is the midpoint of $CD$, we have:
$$ x(c+x) = y(b+y) \tag{2}$$
but we also have that the circumcenter of $FDEG$ is the midpoint of $GE$. 
Let $M$ be such a point and $\Gamma$ be the circumcircle of $FDEG$. Then:
$$ \text{pow}_\Gamma(A)=AM^2-\frac{1}{4}GE^2=\frac{AG^2 + AE^2-GE^2}{2}\tag{3} $$
hence:
$$ x(c+x)=y(b+y)=2\,AG\,AE\,\cos\widehat{A}=-\frac{1}{2}(c+x)(b+y)\cos(\alpha+\beta)\tag{4} $$
from which it is possible fo find $x,y$ in terms of $b,c$ and $\cos(\alpha+\beta)$. Ultimately, that leads to:

$$ BF = \frac{3\cos\alpha-\cos(\alpha+2\beta)}{4-\cos^2(\alpha+\beta)},\qquad CD = \frac{3\cos\beta-\cos(\beta+2\alpha)}{4-\cos^2(\alpha+\beta)}.\tag{5}$$

Now a solution with straightedge and compass: let $J$ be the point such that $JA\perp AB$ and the line through $J$ and the midpoint of $AB$ is orthogonal to $AC$. Let $K$ be the point such that $KA\perp AC$ and the line through $K$ and the midpoint of $AC$ is orthogonal to $AB$. Then, by homothety, the intersection of $FE$ and $DG$ must lie both on $BJ$ and $CK$, hence $FE\cap DG = AJ\cap BK$.

A: 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 $|\overline{CD}| =2z$):

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

A: You can solve this problem by "reverse engineering" it, because for a given triangle $ABC$ it is hard to construct the location of points $F$ and $D$. If you start with the point $A$ though which two lines, with relative angle $0<\gamma<\frac{\pi}{2}$, are drawn and place a point on each of these lines, then you can construct such a triangle, which can be seen in the following figure.

So for now it is assumed that $\gamma$, $AD$ and $AF$ are given. Adding perpendicular lines to the points $D$ and $F$ creates three new intersection points: $E$, $G$ and $H$. From these points, six in total, you can construct 4 similar triangles: $FAE$, $DHE$, $DAG$ and $FHG$. In order to construct the triangle $ABC$ the lengths $DE$ and $FG$ have to be found as a function of $\gamma$, $AD$ and $AF$. There are multiple ways to find this, but the shortest way is by projecting the points $D$ and $E$ perpendicular onto $AF$ and the points $F$ and $G$ perpendicular onto $AD$, such that,
$$
DE = \frac{AF - AD \cos\gamma}{\cos\gamma},
$$
$$
FG = \frac{AD - AF \cos\gamma}{\cos\gamma}.
$$
The lengths of $AB$ and $AC$ can now be expressed in $\gamma$, $DE$ and $FG$,
$$
AB = AF + 2 FG = \frac{DE + FG (\cos\gamma + 2 \sin\gamma \tan\gamma)}{\sin\gamma \tan\gamma},
$$
$$
AC = AD + 2 DE = \frac{FG + DE (\cos\gamma + 2 \sin\gamma \tan\gamma)}{\sin\gamma \tan\gamma}.
$$
These lengths can then be used in combination with the sine law, also using that $BC$ is defined to have length one,
$$
\frac{AB}{\sin\beta} = \frac{AC}{\sin\alpha} = \frac{1}{\sin\gamma}.
$$
By equating these expressions for $AB$ and $AC$, then $DE$ and $FG$ can be expressed as,
$$
DE = \frac{\sin\alpha (\cos\gamma + 2 \sin\gamma \tan\gamma) - \sin\beta}{\sin\gamma \cos\gamma (4 \tan^2\gamma + 3)},
$$
$$
FG = \frac{\sin\beta (\cos\gamma + 2 \sin\gamma \tan\gamma) - \sin\alpha}{\sin\gamma \cos\gamma (4 \tan^2\gamma + 3)}.
$$
The angle $\gamma$ might be substitute by $\pi-\alpha-\beta$. After substitution and some simplification $DE$ and $FG$ can also be expressed as,
$$
DE = \frac{4 \sin\!\left(\alpha\right) - \sin\!\left(3\, \alpha + 2 \beta\right) + 3 \sin\!\left(\alpha + 2 \beta\right)}{15 \sin\!\left(\alpha + \beta\right) - \sin\!\left(3 \alpha + 3 \beta\right)},
$$
$$
FG = \frac{4 \sin\!\left(\beta\right) - \sin\!\left(2 \alpha + 3 \beta\right) + 3 \sin\!\left(2 \alpha + \beta\right)}{15 \sin\!\left(\alpha + \beta\right) - \sin\!\left(3 \alpha + 3 \beta\right)}.
$$
