Challenge: find angle x in the figure via logical approach Background: this problem was from social app, but I cannot find original source. This is an interesting problem that I have not seen similar ones. It is a hard problem, cannot be solved with angle chasing.  I worked out via analytic geometry programming, but cannot figure out a logical approach. Is there a logical approach to solve it? May need help from some theorems that I am not familiar with.
Problem: There are 5 known angles in the figure. Find the angle x. Is there an approach to solve it through deduction?


Label the bottom two points (starting from left) with $A$, $B$, and the top points (starting from left) with $C$, $D$, $E$. Let $F$ be the intersection of $AD$ and $BC$. Let $\omega$ be the circumcircle of $ABF$. Let $G$ be the second common point of $\omega$ and $AE$. Let $H$ be the second common point of $BD$ and $\omega$. Finally, let $I$ be the point on the ray $BA^\to$ such that $AG=AI$.
Since $AI=AG$, we have $\angle AIG = \frac 12 \angle BAG = \frac 12 \cdot 44^\circ = 22^\circ$. Then we find that $\angle GBI = \angle FBI + \angle GBF = 49^\circ + \angle GAF = 49^\circ + 30^\circ = 79^\circ$, and $\angle IGB = 180^\circ - \angle BIG - \angle GBI = 180^\circ - 22^\circ - 79^\circ = 79^\circ$. Hence $IBG$ is isosceles and $IB=IG$.
Let $\Omega$ be the circle with center $I$ and radius $IB=IG$. Let $C'\in\Omega$ be such that the chord $BC'$ passes through $F$. Note that $\angle GIC' = 2 \angle GBC' = 2\angle GBF = 2\cdot 30^\circ = 60^\circ$, hence triangle $GIC'$ is equilateral. In particular, $C'I=C'G$. Since $AI=AG$, it follows that $AC'$ bisects the angle $GAI$. Hence $\angle GAC' = \frac 12 \angle GAI = \frac 12 \cdot (180^\circ - \angle BAG) = 90^\circ - \frac 12 \cdot 44^\circ = 90^\circ - 22^\circ = 68^\circ = \angle GAC$, and therefore $C'$ lies on the line $AC$.
Moreover, $\angle C'BI = 90^\circ - \frac 12 \angle BIC' = 90^\circ - \frac 12(\angle BIG + \angle GIC') = 90^\circ - \frac 12 \cdot (22^\circ + 60^\circ) = 49^\circ = \angle CBI$, and therefore $C'$ lies on the line $BC$.
Hence $C'$ is the intersection of $AC$ and $BC$, hence $C'$ coincides with $C$.
Next, we shall prove that $C, G, H$ are collinear. Indeed, $\angle CGH = \angle CGA + \angle AGH = \angle AIC + (180^\circ - \angle HBA) = \angle GIC + \angle BIG + 180^\circ - 82^\circ = 60^\circ + 22^\circ + 180^\circ - 82^\circ = 180^\circ$.
Applying Pascal theorem for (degenerated) hexagon $FBBHGA$ we see that $BE$ is tangent to $\omega$. Therefore $\angle EBD = \angle BAH = \angle BAF - \angle HAF = 74^\circ - \angle HBF = 74^\circ - 33^\circ = 41^\circ$.


Remark: This is a special case of the following phenomenon:

The proof presented above can be adapted to the general case.
