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...??
 A: Either you have incorrect answer, or you marked the wrong angle. The angle of $x$ should be $20^\circ$.
Hint: Put point $E$  on $AC$ so $AH\perp DE$ and prove that $\triangle HDE$ is equilateral.
A: $\angle CDH = x$ cannot possibly be $40^\circ$.  That would imply $\angle HDB = \angle CDB - \angle CDH = 60^\circ - 40^\circ = 20^\circ$, which would in turn imply that $DH || AC$, hence $D$ is the midpoint of $AB$.  But since $\angle ACD = 40^\circ > 30^\circ = \angle CDB$, this is impossible, since the angle bisector of $\angle ACB$ cuts $AB$ at some point $F$ such that $AF > BF$, which would mean $D$ is between $A$ and $F$, contradicting $\angle ACD > \angle CDB$.
The correct value for $x$ is $20^\circ$.  I do not have an elementary geometric solution, but a trigonometric solution is possible.
A: 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 $F =DE \cap AH.$
$\angle DPG = 30^\circ$
$\therefore \angle DGP = 180^\circ - 100^\circ-30^\circ = 50^\circ \implies \angle FDG = 40^\circ$
$\therefore \boxed{\color{red}x = 60^\circ-40^\circ = 20^\circ}$
