<s>[INST] Find $x$ in the figure 
Find $x$ in the figure.

(Answer: $20^\circ$)

My progress:
Let $P$ such that $PDHC$ is cyclic.

$\angle ACH = 180^\circ-20^\circ-90^\circ = 70^\circ\implies AHC = 100^\circ$
$\therefore ∠PDC=∠AHC=100^\circ$ and $∠PDA=180^\circ-60^\circ-100^\circ=20^\circ$
$\therefore ∠BDH+x=60^\circ$
If I find out $\angle DCP$, ends...??
 [/INST] Leveraging colleague Vasily's perpendicular idea:
Let $P$ be a point on $AH$ such that the quadrilateral $PDHC$ is cyclic.

$\angle ACH = 180^\circ-20^\circ-90^\circ = 70^\circ\implies AHC = 100^\circ.$
$\therefore  ∠PDC=∠AHC=100^\circ$ and $∠PDA=180^\circ-60^\circ-100^\circ=20^\circ.$
Trace $DE \perp AH ~ (E \in AC).$
$G=DC \cap AH$ and $F =DE \cap AH.$
$\angle DPG = 30^\circ$
$\therefore \angle DGP = 180^\circ - 100^\circ-30^\circ = 50^\circ \implies \angle FDG = 40^\circ$
$\therefore \boxed{\color{red}x = 60^\circ-40^\circ = 20^\circ}$
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