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


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:
$$
x:=\widehat{CDE}=\widehat{CAE}=\frac 12\hat A\ .
$$
$\square$
