Find x in the figure (Answer: 40 degrees)
I found the following mathematical relationships but some are still missing.
$\triangle ABC: 80^o+4\theta+2\alpha = 180^o \rightarrow \boxed{2\theta+\alpha=50^o}(i)\\
\triangle CAH:40^o+4\theta+m = 180^o\rightarrow \boxed{m=140^o - 4\theta}(II)\\
\triangle CBG: n+4\theta+\alpha=180^o\rightarrow \boxed{n = 140+ \alpha}(III)\\
\triangle DJE:180^o-40^o-\theta + x + n = 180^o\rightarrow \boxed{x = \theta-n+40^o(IV)}$

 A: Let us compute explicitly $\eta$ in terms of the angles of the given triangle,
$\hat A$, $\hat B$, $\hat C$. They determine all other angles. (One of them is redundant.) So we try to not use $\theta$ as $\theta$, but rather $\frac 14\hat C$, and forget about the notation $\theta$. (Which is used in the figure to replace the two angle bisector constructions.)

We let $\hat A$ be a general angle, it does not need to be $40^\circ$ as in the picture. Then:
$$
\begin{aligned}
\widehat{EFB} &=\widehat{FAB} + \widehat{FBA}=\frac 12\hat A+\frac 12\hat B
=\frac 12(\hat A+\hat B)\ .
\\
2\eta &=90^\circ-\widehat{EFB} 
=\frac 12(180^\circ-(\hat A+\hat B))=\frac 12\hat C\ ,\qquad\text{ so}
\\
\widehat{AED}=\eta &=\frac 12(2\eta)=\frac 14\hat C=\widehat{ACD}
\ ,
\end{aligned}
$$
showing that $ACED$ is (in general) cyclic, and we have in particular:
$$
x:=\widehat{CDE}=\widehat{CAE}=\frac 12\hat A\ .
$$
$\square$
A: $\angle BFE = 90^0 - 2 n = \angle BAF + \angle ABF = 40^0 + \alpha $
So, $\alpha + 2n = 50^0$  ...$(i)$
Now $\angle AFC = 140^0 - 2\theta$, $\angle BFC = 180^0 - \alpha - 2\theta \ $ and
$\angle AFB = 140^0 - \alpha$
But they should add to $360^0$,
So we get $100^0 = 4 \theta + 2\alpha \ $ or $\alpha + 2 \theta = 50^0$ ...$(ii)$
From $(i)$ and $(ii)$,
$\theta = n$
So $\angle ACD = \angle AED$ but as they are both on segment $AD$, $ADEC$ must be cyclic.
Then, $\angle CDE = x = \angle CAE = 40^0$
A: For every angle $\alpha < 50^\circ$ we can draw a figure that matches the givens, by drawing $\triangle ABC$, bisecting the angle at $C$ two times, extending the bisector at $A$ until an arbitrary $E$ where we drop a perpendicular towards $BF$ and bisecting the angle at $E$, etc. So the angles are not determined by the figure, and we'll need to carry a parameter around and hope it cancels out at the end; let's choose $\alpha$ as the parameter.
Without changing any angles, we can slide $E$ farther along $AF$ until the perpendicular towards $BF$ hits it exactly at $B$. Then, by angle sums:
$$ \triangle BCA:\qquad 2\alpha + 4\theta + 80 = 180$$
$$ \triangle BEA:\qquad (\alpha + 90) + 2\eta + 40 = 180 $$
These relations can be solved for $\theta$ and $\eta$ to get
$$ \theta = 25 - \alpha/2 \qquad \qquad \eta = 25 - \alpha/2 $$
Now, counting measuring directions counterclockwise from due right on the diagram,

*

*The direction from $A$ towards $C$ is $80$.

*The direction from $D$ towards $C$ is $80+\theta$

*The direction from $A$ towards $E$ is $40$.

*The direction from $D$ towards $E$ is $40+\eta$.

By subtracting (4) from (2) and recalling from above that $\theta=\eta$, we get $x=40$.
