It suffices to show that $\overline{EF}$ passes through a fixed point $X$ on $(ABC)$. I claim this is the point such that $AX=AC$. We use moving points, animating $D$ linearly, so $F$ is clearly linear as well. $E$ can be found as follows: intersect $(ABC)$ with the reflection of $(ABD)$ about $\overline{BC}$. An inversion (with fixed radius) at $B$ sends $(ABC)$ to a fixed line and $(ABD)$ to a circle through a fixed point and a degree $1$ point, so the image of $E$ has degree $1$. Thus $E$ has degree $2$. Furthermore since $E,X$ lie on the same fixed circle it's well-known that $\overline{EX}$ has degree $1$, so the statement $F \in \overline{EX}$ has degree $2$. Thus we check three cases: When $D=C$, $E,F$ conincide. When $\overline{AD} \perp \overline{AB}$, the projection of $A$ through $D$ onto $(ABC)$ is just the $B$-antipode, so $E=A$. Since $F=A$ as well, $E,F$ again coincide. When $D=\infty_{\overline{BC}}$, we want to show $E$ is the point such that $ACXE$ is an isosceles trapezoid. With a suitable phantom point argument and by reflecting everything across the perpendicular bisector of $\overline{BC}$, it's equivalent to show that if $Y$ is the point on $(ABC)$ with $\angle YBC=90^\circ$ then $Y$ is the midpoint of arc $AE'$, where $E' \in (ABC)$ is defined by having $CA=CE'$. This follows from angle chasing.

. Claim: $E'D'=E'B=E'F$ $$\angle E'BD'=\angle CAD'=2\angle COD'=2\angle CE'D'$$$$\angle D'F'B=90^{\circ}-\angle BAC+\angle BAD'=90^{\circ}-\angle CAD'=90^{\circ}-\frac{1}{2} \angle CE'D'=\frac{1}{2}\angle DE'B$$ To finish, $$\angle CE'F'=180^{\circ}-\angle F'E'B=180^{\circ}-2\angle F'D'B=180^{\circ}-2(\angle ACB+\angle BAC-90^{\circ})=2\angle ABC$$

After $\sqrt{bc}$ invert all we need is: New problem wrote: On $\triangle ABC$ with circumcircle $\omega$ let $H_C$ be a point in $\omega$ such that $CH_C \perp AB$. An arbitrary point $D'$ moves in $\omega$ and now we let $D'H_C \cap AB=F'$. Prove that the circumcenter $E'$ of $\triangle BF'D'$ lies in $BC$. $$\angle E'BD'=90-\angle BF'D'=90-(90-\angle BAC+\angle BAD')=\angle BAC-\angle BAD'=\angle CAD'=\angle CBD'$$Is what gives the desire conclusion, now let $X'$ a point in $BC$ such that $AB=AX'$, note that as $\angle BF'E'=\angle ABC=\angle AXB$ we get that $AF'E'X'$ is cyclic, notice that $D',E',F'$ are inverses of $D,E,F$ in the original problem so inverting back we get $E,F,X$ colinear where $X$ is a fixed point (it's $X'$ inverted) so $\angle BFE=180-\angle BAX$ which is fixed thus we are done .

bruh why invert only once Let $H$ be the foot from $D$ to $AB$. $\sqrt{ac}$ invert. $D'$ is a point on $AB$. $H$ is sent to the point on $BC$ such $\angle BD'H'=90^\circ$. $F'$ is simply the intersection of $(BD'H')$ and $(ABC)$. $E'$ is the intersection of the reflection of $CD'$ over $AB$ with $AC$. Our goal is to prove that $\angle BF'E'$ is fixed. We can just subtract/add fixed angles and it is clear that we only want $\angle AF'E'$ fixed. In fact, it suffices to prove that $AF'$ is the bisector of $\angle CF'E'$, as then $\angle AF'E'=\angle CF'A=\angle CBA$ we would be done. However, clearly $AB$ is the bisector of $\angle CD'E'$. Hence we just want to show that $F'$ lies on the apollonian circle determined by $E',A,C$. Now invert again!!! $\sqrt{bc}$ invert. $D''$ is a point on $AC$. $H''$ is the intersection of semicircle $D''A$ with $(ABC)$. $(D''H''C)$ intersects $BC$ at $F''$. We want to prove that $D''F''$ is perpendicular to $AB$. However, this is a clear angle chase as $\angle CF''D''=90^\circ-\angle CH''A=90^\circ-\angle B$ hence the perpendicularity is clear to see.

This proof is not by me. I'm sending it just because it's supremely elegant, and no one is talking about it here yet. Let $H$ be the orthocenter of $\triangle ABD$. $\angle BED = \angle BAD = \pi - \angle BHD \Rightarrow BHDE$ is cyclic. Hence $\angle FAE = \angle CBE = \angle DHE \Rightarrow AHEF$ is cyclic. Therefore, $\angle BEF = \angle AEB + \angle AEF = \angle ACB + \angle AHD = \angle ACB + \pi - \angle ABC$ is a constant. $\square$