We present several solutions.

First solution (Andrew Gu) We begin with the following two observations.

Claim — Point $K$ lies on the radical axis of $(BEH)$ and $(CFH)$.
Proof. Actually we claim $\overline{KE}$ and $\overline{KF}$ are tangents. Indeed,
$$\angle HEK = 90^\circ - \angle EAF = 90^\circ - \angle BAC = \angle HBE$$
implying the result. Since $KE = KF$, this implies the result. $\square$

Claim — The second intersection $M$ of $(BEH)$ and $(CFH)$ lies on $\Gamma$.
Proof. By Miquel's theorem on $\triangle AEF$ with $H \in \overline{EF}$, $B \in \overline{AE}$, $C \in \overline{AF}$. $\square$

In particular, $M, H, K$ are collinear. Let $X$ be on $\Gamma$ with $\overline{DX} \perp \overline{BC}$; we then wish to show $X$ lies on the line $MHK$ we found. This is angle chasing: compute
$$\begin{aligned} \angle XMB &= \angle XDB = 90^\circ - \angle DBC = 90^\circ - \angle DAC \\ &= 90^\circ - \angle KAF = \angle FEA = \angle HEB = \angle HMB \end{aligned}$$
as needed.

Second solution (Ankan Bhattacharya) We let $D'$ be the second intersection of $\overline{EF}$ with (BHC) and redefine D as the reflection of $D'$ across $\overline{BC}$. We will first prove that this point D coincides with the point D given in the problem statement. The idea is that:

Claim — A is the D-excenter of $\triangle DEF$.
Proof. We contend $BED'D$ is cyclic. This follows by angle chasing:
$$\begin{aligned} \angle D'DB &= \angle BD'D = \angle D'BC + 90^\circ = \angle D'HC + 90^\circ \\ &= \angle D'HC + \angle (HC, AB) = \angle (D'H, AB) = \angle D'EB. \end{aligned}$$
Now as $BD = BD'$, we obtain $\overline{BEA}$ externally bisects $\angle DED' \cong \angle DEF$. Likewise $\overline{FA}$ externally bisects $\angle DFE$, so A is the D-excenter of $\triangle DEF$. $\square$

Hence, by the so-called “Fact 5”, point K lies on $\overline{DA}$, so this point D is the one given in the problem statement.

Now choose point X on (ABC) satisfying $\overline{DX} \perp \overline{BC}$.

Claim — Point K lies on line HX.
Proof. Clearly $AHD'X$ is a parallelogram. By Ptolemy on $DEKF$,
$$\frac{KD}{KA} = \frac{KD}{KE} = \frac{DE + DF}{EF}.$$
On the other hand, if we let $r_D$ denote the D-exradius of $\triangle DEF$ then
$$\frac{XD}{XD'} = \frac{[DEX] + [DFX]}{[XEF]} = \frac{[DEX] + [DFX]}{[AEF]} = \frac{DE \cdot r_D + DF \cdot r_D}{EF \cdot r_D} = \frac{DE + DF}{EF}.$$
Thus
$$[AKX] = \frac{KA}{KD} \cdot [DKX] = \frac{KA}{KD} \cdot \frac{XD}{XD'} \cdot [KD'X] = [D'KX].$$
This is sufficient to prove K lies on $\overline{HX}$. $\square$

The solution is complete: X is the desired concurrency point.

Third solution (Nikolai Beluhov, unedited) We are going to prove the following:
Let $ABC$ be a triangle with orthocenter $H$ and circumcircle $\Gamma$. Let $D$ be any point on arc $BC$ of $\Gamma$ that does not contain $A$. Let $J$ lie on $\Gamma$ so that line $DJ$ is perpendicular to $BC$. Let lines $AD$ and $HJ$ meet at $K$. Let $L$ be such that $K$ is the midpoint of segment $AL$. Let $E$ and $F$ be the projections of $L$ onto lines $AB$ and $AC$, respectively. Then $H$ lies on line $EF$.
This is the converse of the problem statement; clearly, if we prove this, then all is well. Let lines $BC$ and $DJ$ meet at $M$.

Claim — Point $L$ lies on $HM$.
Proof. Let $G$ be the midpoint of segment $AH$, let $O$ be the circumcenter of triangle $ABC$, and let $N$ be the projection of $O$ onto line $DJ$. Then $N$ is the midpoint of segment $DJ$, so $K$ lies on $GN$. Also $GH$ equals the distance from $O$ to line $BC$, which equals $MN$; thus $GN$ is parallel to $HM$. It follows that $GK$ is parallel to both $HL$ and $HM$. $\square$

Let $P$ and $Q$ be the projections of $D$ onto lines $AB$ and $AC$, respectively. Let $\ell$, the line through $P, M, Q$ be the Simson line of $D$ with respect to triangle $ABC$. Suppose $\ell$ meets line $AH$ at $R$.

Claim — We have $DM = HR$.
Proof. This is a known property of the Simson line $\ell$ (that $DMHR$ is in fact a parallelogram as $\ell$ bisects $\overline{HD}$). $\square$

Claim — Figures $ALEFH$ and $ADPQR$ are homothetic with center $A$.
Proof. All that we need to do to establish this is to verify that $AL : LD = AH : HR$. This is true by $AL : LD = AH : DM = AH : HR$. $\square$

By the final claim, since $R$ lies on line $PQ$, we get that $H$ lies on line $EF$. This completes the solution.

Fourth solution, complex numbers with spiral similarity (Evan Chen) First if $\overline{AD} \perp \overline{BC}$ there is nothing to prove, so we assume this is not the case. Let $W$ be the antipode of $D$. Let $S$ denote the second intersection of $(AEF)$ and $(ABC)$. Consider the spiral similarity sending $\triangle SEF$ to $\triangle SBC$:
* It maps $H$ to a point $G$ on line $BC$,
* It maps $K$ to $O$.
* It maps the $A$-antipode of $\triangle AEF$ to $D$.
* Hence (by previous two observations) it maps $A$ to $W$.
* Also, the image of line $AD$ is line $WO$, which does not coincide with line $BC$ (as $O$ does not lie on line $BC$).

Therefore, $K$ is the unique point on line $\overline{AD}$ for one can get a direct similarity
$$\triangle AKH \sim \triangle WOG \quad (\heartsuit)$$
for some point $G$ lying on line $\overline{BC}$.

On the other hand, let us re-define $K$ as $\overline{XH} \cap \overline{AD}$. We will show that the corresponding $G$ making $(\heartsuit)$ true lies on line $BC$.
We apply complex numbers with $\Gamma$ the unit circle, with $a, b, c, d$ taking their usual meanings, $H = a+b+c$, $X = -bc/d$, and $W = -d$. Then point $K$ is supposed to satisfy
$$\begin{aligned} k + a\bar{d}k &= a + d \\ \frac{k + \frac{bc}{d}}{a + b + c + \frac{bc}{d}} &= \frac{\bar{k} + \frac{d}{bc}}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc}} \\ \iff \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc}}{a + b + c + \frac{bc}{d}} ( k + \frac{bc}{d} ) &= \bar{k} + \frac{d}{bc} \end{aligned}$$
Adding $ad$ times the last line to the first line and cancelling $a\bar{d}k$ now gives
$$\left( a d \cdot \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc}}{a + b + c + \frac{bc}{d}} + 1 \right) k = a + d + \frac{a d^2}{bc} - a b c \cdot \frac{\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc}}{a + b + c + \frac{bc}{d}}$$
or
$$\left( a d \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc} \right) + a + b + c + \frac{bc}{d} \right) k = \left( a + b + c + \frac{bc}{d} \right) \left( a + d + \frac{a d^2}{bc} \right) - a b c \cdot \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc} \right).$$
We begin by simplifying the coefficient of $k$:
$$\begin{aligned} a d \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{d}{bc} \right) + a + b + c + \frac{bc}{d} &= a + b + c + d + \frac{bc}{d} + \frac{ad}{b} + \frac{ad}{c} + \frac{ad^2}{bc} \\ &= a + \frac{bc}{d} + \left( 1 + \frac{ad}{bc} \right) (b + c + d) \\ &= \frac{ad + bc}{bcd} [bc + d(b + c + d)] \\ &= \frac{(ad + bc)(d + b)(d + c)}{bcd}. \end{aligned}$$

Meanwhile, the right-hand side expands to
$$\begin{align*} \text{RHS} &= \left(a+b+c+\frac{bc}{d}\right)\left(a+d+\frac{ad^2}{bc}\right) - abc \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{d}{bc}\right) \\ &= \left(a^2+ab+ac+\frac{abc}{d}\right) + (da+db+dc+bc) \\ &\quad + \left(\frac{a^2d^2}{bc}+\frac{ad^2}{c}+\frac{ad^2}{b}+ad\right) - (ab+bc+ca+ad) \\ &= a^2+d(a+b+c) + \frac{abc}{d} + \frac{a^2d^2}{bc} + \frac{ad^2}{b} + \frac{ad^2}{c} \\ &= a^2 + \frac{abc}{d} + d(a+b+c) \cdot \frac{ad+bc}{bc} \\ &= \frac{ad+bc}{bcd} \left[abc+d^2(a+b+c)\right]. \end{align*}$$
Therefore, we get
$$k = \frac{abc + d^2(a + b + c)}{(d + b)(d + c)}.$$
In particular,
$$\begin{align*} k - a &= \frac{abc + d^2(a + b + c) - a(d + b)(d + c)}{(d + b)(d + c)} \\ &= \frac{d^2(b + c) - da(b + c)}{(d + b)(d + c)} = \frac{d(b + c)(d - a)}{(d + b)(d + c)}. \end{align*}$$
Now the corresponding point $G$ obeying $(\heartsuit)$ satisfies
$$\begin{align*} \frac{g - (-d)}{0 - (-d)} &= \frac{(a + b + c) - a}{k - a} \\ \implies g &= -d + \frac{d(b + c)}{k - a} \\ &= -d + \frac{(d + b)(d + c)}{d - a} = \frac{db + dc + bc + ad}{d - a}. \\ \implies bc\bar{g} &= \frac{bc \cdot \frac{ac+ab+ad+bc}{abcd}}{\frac{a-d}{ad}} = -\frac{ab + ac + ad + bc}{d - a}. \\ \implies g + bc\bar{g} &= \frac{(d - a)(b + c)}{d - a} = b + c. \end{align*}$$
Hence $G$ lies on $BC$ and this completes the proof.

Fifth solution by trigonometry (Ivan Borsenco, unedited) Let $\angle E = B - \theta$ and $\angle F = C - \theta$. Denote by $H_a$ the intersection of $AH$ with $\Gamma$ and by $D'$ the intersection of the line passing through $D$ and perpendicular to $BC$ with $EF$.
By angle-chasing, we get $\angle BAH = 90^\circ - B$, $\angle AHE = \angle H_aHD' = 90^\circ + \theta$. On the other hand, $\angle ADH_a = \angle ACH_a = 90^\circ - B + C$, $\angle AKB = 2(C+\theta)$, $\angle H_aAK = \angle HAK = \angle EAK - \angle EAH = 90^\circ - (C+\theta) - (90^\circ - B) = B - C - \theta$, and therefore $\angle AH_aD = 90^\circ + \theta$. Hence $HH_aDD'$ is an isosceles trapezoid. Point $D'$ is the reflection of $D$ in $BC$, which implies that quadrilateral $BHD'C$ is cyclic, because $\angle BHC = \angle BD'C = 180^\circ - A$.
Choose point $X$ on $\Gamma$ satisfying $DX \perp BC$. Note that $AH_aDX$ is an isosceles trapezoid. Hence $\angle HAX = 90^\circ + \theta$ and $\angle KAX = \angle HAX - \angle HAK = 90^\circ - (B-C-2\theta)$. Denote

by $R$ and $R'$ the circumradii of triangles $ABC$ and $AEF$, respectively. It follows that $AH = 2R \cos A$, $AK = R'$, $AX = DH_a = 2R \sin(B - C - \theta)$.
In order to show that points $H$, $K$, and $X$ are collinear, we will show that $[HAX] = [HAK] + [KAX]$, which is equivalent to
$$AH \cdot AX \cdot \sin(\angle HAX) = AK [AH \cdot \sin(\angle HAK) + AX \cdot \sin(\angle KAX)].$$
Using the Law of Sines in triangles $AEH$ and $AFH$, we get
$$EF = EH + HF = AH \cdot \frac{\cos B}{\sin(B - \theta)} + AH \cdot \frac{\cos C}{\sin(C + \theta)}.$$
yielding
$$\begin{aligned} 2R' \sin A &= \frac{2R \cos A}{\sin(B - \theta) \sin(C + \theta)} (\cos B \sin(C + \theta) + \cos C \sin(B - \theta)) \\ &= \frac{2R \cos A}{\sin(B - \theta) \sin(C + \theta)} \cdot \cos \theta \cdot \sin(B + C). \end{aligned}$$
Using the fact that $2 \sin(B - \theta) \sin(C + \theta) = \cos((B - \theta) - (C + \theta)) - \cos((B - \theta) + (C + \theta))$, we conclude that
$$R' = 2R \cdot \frac{\cos A \cos \theta}{\cos(B - C - 2\theta) + \cos A}.$$
Denote by $\varphi = B - C - 2\theta$, then
$$AH = 2R \cos A, \quad AK = 2R \cdot \frac{\cos A \cos \theta}{\cos \varphi + \cos A}, \quad AX = 2R \sin(\varphi + \theta),$$
and
$$\angle HAK = \varphi + \theta, \quad \angle KAX = 90^\circ - \varphi, \quad \angle HAX = 90^\circ + \theta.$$
Returning back to proving the identity for areas, we have to show that
$$\cos A \cdot \sin(\varphi + \theta) \cdot \cos \theta = \frac{\cos A \cos \theta}{\cos \varphi + \cos A} \cdot [\cos A \cdot \sin(\varphi + \theta) + \sin(\varphi + \theta) \cdot \cos \varphi],$$
which is clearly true.

Sixth solution by moving points (Anant Mudgal, unedited) The meat of this solution is the following claim.

Claim — In triangle $AEF$, with circumcenter $K$ point $H$ lies on $\overline{EF}$, points $B$ and $C$ lie on lines $\overline{AE}$ and $\overline{AF}$ respectively, such that $\overline{BH} \perp \overline{AF}$ and $\overline{CH} \perp \overline{AE}$. Line $\overline{AK}$ meets $\odot(KEF)$ again at point $D$. Then $ABCD$ is cyclic and reflection of $D$ in $\overline{BC}$ lies on $\overline{EF}$.
Proof. Move $H$ along $\overline{EF}$ and note that $B \mapsto H$ and $H \mapsto C$ are linear maps, hence $B \mapsto C$ is also linear. Suppose $\odot(DAB)$ meets line $\overline{AF}$ at $C'$. Then we need to show that $C = C'$. Since, by spiral similarity, $B \mapsto C'$ is linear; we need to check this for two choices of $H$.
* $H = E$. Then $B = E$ and we need to show that if $\odot(AED)$ meets $\overline{AF}$ at $F'$, then $\angle AEF' = 90^\circ$. Apply inversion at $A$ of radius $\sqrt{AE \cdot AF}$ followed by reflection in the bisector of angle $EAF$. Suppose $X \mapsto X^*$ under this transformation. Then $E^* = F, F^* = E$ and $D^*$ is the orthocenter of $\triangle AEF$, so $(F')^* = \overline{FD^*} \cap \overline{AE}$ hence $\angle A(F')^*F = 90^\circ$ so $\angle AEF' = 90^\circ$, and we're done.
* $H = F$. Same proof as above works.
Finally, moving $H$, since $\triangle DBC$ has fixed shape, so the locus of the reflection of $D$ in $\overline{BC}$ is a line.
* For $H = E$, we need to show that $\angle AED = 180^\circ - \angle AEF$ since $\angle FEF' = 90^\circ - \angle AEF$; this follows since $\angle AED = \angle AD^*F$ and $D^*$ is the orthocenter of $\triangle AEF$.
* Similarly, $H = F$ case holds.
The lemma is proved. $\square$

Now we go back to the original problem. Let $L$ be the reflection of $A$ in $K$ and $N = \overline{EF} \cap \overline{AK}$, then, by our lemma, we have $(AL; ND) = -1$.
Suppose $P$ lies on $\overline{EF}$ such that $\overline{DP} \perp \overline{BC}$ and $\overline{DP}$ meets $\overline{BC}$ at $S$ and $\Gamma$ again at $Q$. Reflect $Q$ in $S$ to get $R$. By the lemma, $S$ is the midpoint of $\overline{DP}$. Let $S' = \overline{HL} \cap \overline{DP}$ and $Q' = \overline{HK} \cap \overline{DP}$.
Observe that $-1 = (AL; ND) \stackrel{H}{=} (\infty s', PD)$, so clearly, $\overline{HL}$ bisects $\overline{DP}$, so $H, L, S$ are collinear. Finally, since $\overline{AK} \parallel \overline{RH}$ so $-1 = (AL; K\infty) \stackrel{H}{=} (\infty S; Q'R)$ so $\overline{HK}$ passes through $Q$, as desired.

Seventh solution using brutal force (Zack Chroman) We state the converse of the problem as follows:
Take a point $D$ on $\Gamma$, and let $G \in \Gamma$ such that $\overline{DG} \perp \overline{BC}$. Then define $K$ to lie on $\overline{GH}$, $\overline{AD}$, and take $L \in \overline{AD}$ such that $K$ is the midpoint of $\overline{AL}$. Then if we define $E$ and $F$ as the projections of $L$ onto $\overline{AB}$ and $\overline{AC}$ we want to show that $E, H, F$ are collinear.
It's clear that solving this problem will solve the original. In fact we will show later that each line $EF$ through $H$ corresponds bijectively to the point $D$.
We work in the real projective plane $\mathbb{RP}^2$, and animate $D$ on $\Gamma$. The point $D$ has projective coordinates which are each quadratic polynomials in a real parameter $t$, and moves projectively on $(ABC)$. We will state and prove some quick facts about animation. First, define the degree of a moving point $(P(t) : Q(t) : R(t))$ to be the max degree of $P, Q, R$. Similarly we define the degree of a moving line $P(t)x + Q(t)y + R(t)z = 0$ in the same way.

Lemma
Suppose points $A, B$ have degree $d_1, d_2$, and there are $k$ values of $t$ for which $A = B$. Then line $AB$ has degree at most $d_1 + d_2 - k$. Similarly, if lines $l_1, l_2$ have degrees $d_1, d_2$, and there are $k$ values of $t$ for which $l_1 = l_2$, then the intersection $l_1 \cap l_2$ has degree at most $d_1 + d_2 - k$.
Proof. We show the first statement; the second follows from point-line duality. Note that the line through the points $A = (P_1(t) : Q_1(t) : R_1(t))$ and $B = (P_2(t) : Q_2(t) : R_2(t))$ is given by cross product $A \times B$; that is, the line
$$(Q_1R_2 - Q_2R_1)x + (R_1P_2 - R_2P_1)y + (P_1Q_2 - P_2Q_1)z = 0.$$
Clearly $A$ and $B$ lie on this line, so it is line $AB$. Then for every value $t_0$ for which $A = B$, $(t - t_0)$ factors out of each term. So the degree of the line is at most $d_1 + d_2 - k$. $\square$

Now, note that $G$ is projective in $D$ since it's a projection through the point at infinity on line $AH$. Now by the lemma, line $HG$ has degree at most 2, and line $AD$ has degree at most 1.
So by the lemma again, the point $K$ has degree at most 3. However, note that when $D$ lies on line $AH$, we have $G = A$, so lines $HG$ and $AD$ are the same. It follows that the point $K$ actually has degree at most 2, thus so does $L$.
Let $P_C$ be the point at infinity on the line perpendicular to $AC$, and similarly $P_B$. Then
$$F = \overline{AC} \cap \overline{P_C L}, \quad E = \overline{AB} \cap \overline{P_B L},$$
so $E$ and $F$ have degree at most 2, since lines $AB$ and $AC$ are fixed and $\deg(P_B L) \le \deg(P_B) + \deg(L) = 2$. In fact, note that if we can show that $P_B, P_C$ lie on the locus of $L$, we'll show that $E$ and $F$ move with degree 1 (i.e. projectively) by the lemma again. To show that, we consider the case where $L$ and $K$ lie at infinity; that is, $\overline{HG} \parallel \overline{AD}$. In this case, $ADGH$ is a parallelogram as $AH \parallel DG$. Clearly $G = B$ and $G = C$ work; when $G = B$, $D$ is the antipode of $C$ in $\Gamma$.
Then, when $G = B$, we have $K = L$ is the point at infinity on line $\overline{GH} \equiv \overline{BH}$. This point is $P_C$, so we get that $E, F$ are projective.
So it suffices to verify the problem for three distinct choices of $D$.
* If $D = A$, then line $GH$ is line $AH$, and $L = \overline{AD} \cap \overline{AH} = A$. So $E = F = A$ and the statement is true.
* If $D = B$, $G$ is the antipode of $C$ on $\Gamma$. Then $K = \overline{HG} \cap \overline{AD}$ is the midpoint of $\overline{AB}$, so $L = B$. Then $E = B$ and $F$ is the projection of $B$ onto $AC$, so $E, H, F$ collinear.
* We finish similarly when $D = C$.
Thus since the maps $D \mapsto E$ and $D \mapsto F$ are collinear, the map $E \mapsto F$ is projective as well. Since $E, H, F$ are collinear for three values of $E$, they are in general. Moreover, since $D \to E$ is bijective, any line through $H$ will correspond to some $D$, so we've solved the original problem as well.

Eight solution by author using circumhyperbolas (Gunmay Handa, unedited) Let $P$ be an arbitrary point on $\odot(ABC)$ with $N$ as the midpoint of $\overline{HP}$, and define $\mathcal{H}_P = ABCHP$ as the rectangular circumhyperbola with center $N$ passing through the aforementioned points. Moreover, define $D' \in \odot(ABC)$ with $\overline{PD'} \perp \overline{BC}$ and $P \neq D'$; observe that the line $\ell_P$ through $O$ perpendicular to $\overline{AD'}$ is the isogonal conjugate of $\mathcal{H}_P$ with respect to $\odot(ABC)$, and so if we define $U$ and $V$ as the intersections of $\ell_P$ with $\overline{AB}$ and $\overline{AC}$, respectively, then $N$ belongs to the pedal circles $\omega_U$ and $\omega_V$ of $U$ and $V$ with respect to $\triangle ABC$.
Let $\triangle RST$ be the orthic triangle of $\triangle ABC$ and $M$ be the midpoint of $\overline{AH}$; angle chasing establishes that if $\{Q, N\} = \omega_U \cap \omega_V$, then $Q \in \odot(ABC)$, and moreover $H \in \overline{QN}$ since it has equal power with respect to these circles. Suppose the line through $H$ parallel to $\overline{AP}$ intersects $\overline{AB}$ and $\overline{AC}$ at $E'$ and $F'$, respectively, and observe that $\overline{E'F'}$ is antiparallel to $\overline{UV}$ in $\angle A$. If $K'$ is the orthocenter of $\triangle AUV$, then $K' \in \overline{QN}$ by radical axes, and moreover $K' \in \odot(UD'V)$ since $D'$ is the reflection of $A$ across $\overline{UV}$. Further angle chasing establishes $Q \in \odot(UD'V)$; we now claim that $E', F' \in \odot(UD'V)$ as well. Suppose the line through $N$ parallel to $\overline{AP}$ intersects $\overline{AB}$ at $W$, so that since $\triangle AST \cup \overline{MW} \sim \triangle ABC \cup \overline{OV}$, we have that $AK' \cdot AD'/2 = AW \cdot AU = AE'/2 \cdot AU$, and so $E', F' \in \odot(UD'V)$ as well. Finally, since $\angle EUK = \angle FVK = 90^\circ - \angle A$, we know

that $\overline{DK}$ bisects $\angle EDF$, which implies that $K'$ is the circumcenter of $\odot(AE'F')$ since $\overline{AK'} \perp \overline{UV}$ and lines $E'F'$ and $UV$ are isogonal in $\angle A$, which finishes the problem.