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

 [/INST] $\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, $\angle CDE = x = \angle CAE = 40^0$
</s>