Find $x$ in the figure

Question

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...??

Answer 1

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 $F =DE \cap AH.$

$\angle DPG = 30^\circ$

$\therefore \angle DGP = 180^\circ - 100^\circ-30^\circ = 50^\circ \implies \angle FDG = 40^\circ$

$\therefore \boxed{\color{red}x = 60^\circ-40^\circ = 20^\circ}$

Answer 2

$\angle CDH = x$ 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.

Answer 3

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.