Let $T$ be the point on $\Gamma$ such that $\overline{DT}\perp\overline{BC}$, and let $Q = \overline{TH}\cap \Gamma$. [asy][asy] defaultpen(fontsize(10pt)); size(250); pair A, B, C, H, E, F, K, D, Q, T, X; A = dir(117); B = dir(210); C = dir(330); H = orthocenter(A, B, C); X = dir(20); E = extension(H, X, A, B); F = extension(H, X, A, C); K = circumcenter(A, E, F); D = IP(Line(A, K, 20), circumcircle(A, B, C), 1); T = IP(Line(D, foot(D, B, C), 20), circumcircle(A, B, C), 1); Q = IP(Line(T, H, 20), circumcircle(A, B, C), 1); draw(A--B--C--cycle, orange); draw(circumcircle(A, B, C), red); draw(E--F, red); draw(circumcircle(A, E, F), heavygreen); draw(A--D, orange); draw(circumcircle(B, E, H), purple); draw(circumcircle(C, F, H), purple); draw(Q--T, magenta); draw(E--K--F, magenta+linewidth(0.9)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(180)); dot("$F$", F, dir(35)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$H$", H, dir(290)); dot("$Q$", Q, dir(210)); dot("$K$", K, dir(30)); [/asy][/asy] Claim: $BEHQ$ and $CFHQ$ cyclic. Proof. Pure angles: $$\angle BQH = \angle BDT = 90^\circ - \angle DBC = 90^\circ - \angle KAF = \angle AEF.$$A similar angle chase proves $CFHQ$ cyclic. $\blacksquare$ Now $\overline{QHT}$ is the radical axis of $(BEHQ)$ and $(CFHQ)$. To finish, observe that $\angle KEH = 90^\circ - \angle A = \angle EBH$, which implies $\overline{KE}$ tangent to $(BEHQ)$; similarly, $\overline{KF}$ is tangent to $(CFHQ)$. Thus, since $KE = KF$, $K$ lies on the radical axis, as desired.

Let $X$ be the point of $\odot(ABC)$ such that $XD\perp BC$. First, by angle chasing $$\angle XBC = 90^{\circ} - \angle BXD = 90^{\circ} - \angle EAK = \angle AFE$$so $\triangle XBC\cup O\sim\triangle AFE\cup K$. Moreover, if $A'$ be the antipode of $A$ w.r.t. $\odot(ABC)$ and $XA'$ meet $BC$ at $P$. Then $$\angle BXP = 90^{\circ}-\angle C = \angle FAH$$so $\triangle XBC\cup O\cup P\sim\triangle AFE\cup K\cup H$ which means $\angle AHK = \angle OPX$. Hence we will be done if we prove that $\angle AHX = \angle OPX$. Let $M,N$ be midpoints of $BC, XA'$. Notice that the homothety $\mathcal{H}(A',2)$ takes $\triangle OMN$ to $\triangle AHX$. Moreover, $O, P, M, N$ are concyclic. Thus $\angle AHX = \angle OMN = \angle OPN$ so we are done.

I also like moving points

wait I like this problem why do so many of the TSTST problems this year have orthocenters in their geo

Hmm I guess since i've been called out...

$\measuredangle$ denotes a directed angle modulo $180^\circ$. Denote by $H_A$, $H_B$, $H_C$ the reflections of $H$ over $\overline{BC}$, $\overline{CA}$, $\overline{AB}$, respectively. [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O,A,B,C,H,K,EE,F,D,T,SS,HA,HB,HC; O=(0,0); A=dir(110); B=dir(210); C=dir(330); H=A+B+C; EE=(2A+5B)/7; F=extension(EE,H,A,C); K=circumcenter(A,EE,F); D=2*foot(O,A,K)-A; T=intersectionpoint(H -- (H+(K-H)*100),circle(O,1)); SS=2*foot(O,H,K)-T; HA=2*foot(O,A,H)-A; HB=2*foot(O,B,H)-B; HC=2*foot(O,C,H)-C; filldraw(circle(O,1),fil,pri); filldraw(circumcircle(A,EE,F),sfil,sec); filldraw(circumcircle(D,F,K),sfil,sec); filldraw(circumcircle(B,EE,H),tfil,tri); filldraw(circumcircle(C,F,H),tfil,tri); draw(A--B--C--A,pri); draw(A--D,sec); draw(A--HA,sec); draw(B--HB,sec); draw(C--HC,sec); draw(EE--F,pri); draw(HB--D--HC,tri); draw(SS--T,tri); dot("$A$",A,N); dot("$B$",B,dir(195)); dot("$C$",C,dir(-15)); dot("$H$",H,dir(120)); dot("$K$",K,dir(-40)); dot("$E$",EE,dir(180)); dot("$F$",F,dir(30)); dot("$D$",D,S); dot("$T$",T,T); dot("$S$",SS,SW); dot("$H_A$",HA,NE); dot("$H_B$",HB,HB); dot("$H_C$",HC,HC); [/asy][/asy] Claim 1. $D$ is the anti-Steiner point of $\overline{EHF}$, and $DEKF$ is cyclic. Proof. Let $D'=\overline{EH_C}\cap\overline{FH_B}$ be the anti-Steiner point of $\overline{EHF}$. First note that \begin{align*} \measuredangle KED'&=\measuredangle KEA+\measuredangle AED'=90^\circ-\measuredangle AFE+\measuredangle AEH_C\\ &=90^\circ+\measuredangle EFA+\measuredangle HEA=90^\circ+\measuredangle EFA+\measuredangle FEA, \end{align*}which is symmetric, whence $\measuredangle KED'=90^\circ+\measuredangle EFA+\measuredangle FEA=\measuredangle KFD'$, and $D'EKF$ is cyclic. Now, $$\measuredangle AKE=2\measuredangle AFE=2\measuredangle AFH=\measuredangle H_BFH=\measuredangle D'FE=\measuredangle D'KE,$$so $D'\in\overline{AK}$, as required. $\blacksquare$ Claim 2. Let $S$ be the Miquel point of $EFH_BH_C$; $S$ lies on line $HK$. Proof. Check that $$\frac{SE}{SF}=\frac{EH_C}{FH_B}=\frac{EH}{FH},$$whence $\overline{SH}$ bisects $\angle ESF$ and $\overline{SH}$ passes through $K$, the midpoint of arc $EF$ on $(DEKF)$. $\blacksquare$ Claim 3. Quadrilaterals $BEHS$ and $CFHS$ are cyclic. Proof. Notice that $$\measuredangle ESH=\measuredangle ESK=\measuredangle EFK=90^\circ-\measuredangle FAE=\measuredangle ABH=\measuredangle EBH,$$so we are done by symmetry. $\blacksquare$ Finally, let line $HK$ meet $\Gamma$ at $T\ne S$. Note that $$\measuredangle ATH=\measuredangle ATS=\measuredangle ABS=\measuredangle EBS=\measuredangle EHS=\measuredangle FHT,$$so $\overline{AT}\parallel\overline{EF}$. However, $\overline{DH_A}$ is the reflection of $\overline{EF}$ over $\overline{BC}$, so $\measuredangle TAH_A=\measuredangle EHA=\measuredangle AH_AD$, and $AD=TH_A$. Thus, $ATDH_A$ is an isosceles trapezoid, and $\overline{AH_A}\parallel\overline{DT}$. This completes the proof. $\square$

What is this moving points technique and where can I read more about it?

pinetree1 wrote: Let $ABC$ be an acute triangle with orthocenter $H$ and circumcircle $\Gamma$. A line through $H$ intersects segments $AB$ and $AC$ at $E$ and $F$, respectively. Let $K$ be the circumcenter of $\triangle AEF$, and suppose line $AK$ intersects $\Gamma$ again at a point $D$. Prove that line $HK$ and the line through $D$ perpendicular to $\overline{BC}$ meet on $\Gamma$. Gunmay Handa We first prove the following lemma. Lemma. 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, SO $B \mapsto C$ 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 $-1=(AL; ND) \overset{H}{=} (\infty s', PD)$, so clearly, $\overline{HL}$ bisects $\overline{DP}$, so $H, L, S$ are collinear. Finally, since $\overline{AK} \parallel \overline{RH}$ so $-1=(AL; K \infty) \overset{H}{=} (\infty S; Q'R)$ so $\overline{HK}$ passes through $Q$, as desired.

Define $H_B, H_C$ as the reflections of $H$ over $AC, AB$, respectively and let $D_A$ be the reflection of $D$ over $BC$. Claim: $H_BE$ and $H_CF$ concur at $D$. Proof: By Pascal, they concur at a point $D' \in \Gamma$. Define $A' = AD' \cap \odot(AEF)$. Observe that \begin{align*} \angle AFA' &= \angle EFA' + \angle AFE \\ &=\angle EAA' + (90^{\circ} - \angle FHH_C) \\ &=90^{\circ} + \angle D'AC - \angle D'H_CC \\ &=90^{\circ}, \end{align*}so $AA'$ is a diameter and $D=D'$, as desired. $\Box$ Claim: $D_A \in EF$ Proof: Let $D_B$ be the reflection of $D$ over $AC$. Then $\angle CED_B = \angle DEC = \angle AEH_B = \angle HEA$, so $D_B \in EF$. But it is well-known that the Simson line of $D$ bisects $\overline{DH}$, i.e. $H, D_A, D_B$ collinear. Hence the claim follows. $\Box$ Define $Y = BC \cap DD_A$ and $X = DD' \cap \Gamma$, and then $X'$ as the reflection of $X$ over $\overline{BC}$. Note that $AXD_AH$ is a parallelogram, so $AH = XD_A = DX'$ and $AHX'D$ is also a parallelogram. Now we have \[-1 = (A, A'; D, EF \cap AA') \stackrel{H}{=} (\infty, HA' \cap XD; D, D_A)\]but since $Y$ is the midpoint of $DD_A$ we must have $H, A', Y$ collinear. Finally, we have \[-1=(X,X';Y,\infty) \stackrel{H}{=} (HX \cap AA', \infty; A', A)\]so $HX$ bisects $\overline{AA'}$, which is what we needed to prove. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.5) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -11.1, xmax = 8.62, ymin = -6.63, ymax = 4.59; /* image dimensions */ pen dbwrru = rgb(0.8588235294117647,0.3803921568627451,0.0784313725490196); pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); pen wqwqwq = rgb(0.3764705882352941,0.3764705882352941,0.3764705882352941); /* draw figures */ draw(circle((-4.33,0.13020408163265274), 3.110850339950104), linewidth(1) + dbwrru); draw(circle((-5.476318796333001,1.226069786482878), 1.6641554483504366), linewidth(1) + dbwrru); draw((-6.76760594795539,0.17631226765799224)--(-3.8221458329397446,1.0440665902109125), linewidth(1) + wrwrwr); draw((-4.535744853383504,-2.9738350439058583)--(-7.326646840148698,0.965367271071998), linewidth(1) + wrwrwr); draw((-4.535744853383504,-2.9738350439058583)--(-3.454292693656876,3.115254592291552), linewidth(1) + wrwrwr); draw((-3.454292693656876,3.115254592291552)--(-7.2,-1.07), linewidth(1) + wrwrwr); draw((-7.2,-1.07)--(-5.84,2.85), linewidth(1) + wrwrwr); draw((-5.84,2.85)--(-1.46,-1.07), linewidth(1) + wrwrwr); draw((-1.46,-1.07)--(-7.2,-1.07), linewidth(1) + wqwqwq); draw((-5.84,2.85)--(-5.84,0.4495918367346942), linewidth(1) + wrwrwr); draw((-4.535744853383504,3.234243207171163)--(-4.535744853383503,-5.37424320717117), linewidth(1) + wrwrwr); draw((-4.535744853383503,-5.37424320717117)--(-5.84,0.4495918367346942), linewidth(1) + wrwrwr); draw((-7.326646840148698,0.965367271071998)--(-1.46,-1.07), linewidth(1) + wrwrwr); draw((-5.84,2.85)--(-4.535744853383504,-2.9738350439058583), linewidth(1) + wrwrwr); draw((-5.84,0.4495918367346942)--(-4.535744853383504,-1.07), linewidth(1) + linetype("4 4") + wrwrwr); draw((-5.84,0.4495918367346942)--(-4.535744853383504,3.234243207171163), linewidth(1) + linetype("4 4") + wrwrwr); /* dots and labels */ dot((-5.84,2.85),dotstyle); label("$A$", (-5.96,3.05), NE * labelscalefactor); dot((-7.2,-1.07),dotstyle); label("$B$", (-7.7,-1.21), NE * labelscalefactor); dot((-1.46,-1.07),dotstyle); label("$C$", (-1.38,-1.21), NE * labelscalefactor); dot((-7.326646840148698,0.965367271071998),linewidth(3pt) + dotstyle); label("$H_C$", (-7.96,1.19), NE * labelscalefactor); dot((-3.454292693656876,3.115254592291552),linewidth(3pt) + dotstyle); label("$H_B$", (-3.38,3.27), NE * labelscalefactor); dot((-5.84,0.4495918367346942),linewidth(3pt) + dotstyle); label("$H$", (-6.18,0.59), NE * labelscalefactor); dot((-6.76760594795539,0.17631226765799224),dotstyle); label("$F$", (-7.16,-0.05), NE * labelscalefactor); dot((-3.8221458329397446,1.0440665902109125),linewidth(3pt) + dotstyle); label("$E$", (-3.74,1.21), NE * labelscalefactor); dot((-4.535744853383504,-2.9738350439058583),linewidth(3pt) + dotstyle); label("$D$", (-4.46,-2.81), NE * labelscalefactor); dot((-5.112637592666003,-0.39786042703424335),linewidth(3pt) + dotstyle); label("$A'$", (-5.04,-0.23), NE * labelscalefactor); dot((-4.535744853383504,3.234243207171163),linewidth(3pt) + dotstyle); label("$X$", (-4.46,3.39), NE * labelscalefactor); dot((-4.535744853383504,0.8338350439058582),linewidth(3pt) + dotstyle); label("$D_A$", (-5.1,0.88), NE * labelscalefactor); dot((-4.535744853383503,-5.37424320717117),linewidth(3pt) + dotstyle); label("$X'$", (-4.46,-5.21), NE * labelscalefactor); dot((-4.535744853383504,-1.07),linewidth(3pt) + dotstyle); label("$Y$", (-4.46,-0.91), NE * labelscalefactor); dot((-5.476318796333001,1.2260697864828787),linewidth(3pt) + dotstyle); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Here is another moving points solution based on discussion with yayups. We use zacchro's definition of degree of points. Move point $E$ along $AB$. Then we observe the following. Point $F$ has degree $1$ as $E\mapsto F$ is projective. Let $D'$ be the reflection of $D$ across perpendicular bisector of $BC$. This means $AD'\perp EF$ so $E\mapsto D'\mapsto D$ are projective maps. Midpoints $E'$, $F'$ of $AE, AF$ have degree one. This means that point $K$ is induced by two pencils from $\infty_{\perp AB}$ and $\infty_{\perp AC}$ i.e. $K$ moves along a hyperbola $\mathcal{H}$. As $AD$ has degree $1$, $K$ has degree $2$. Let $X$ be the point on $\odot(ABC)$ such that $XD\perp BC$. Clearly $D\mapsto X$ is projective. In particular, $X$ has degree two. This means that degree of $H,K,X$ are colinear is $0+2+2=4$. Therefore we have to check the problem for five cases. Now, let's check five cases. When $E=B$, we get that $F=H_b$ is the foot of altitude from $B$. Thus $K$ is midpoint of $AB$, $D=B$ and $X$ is the antipode of $C$. The result is then obvious. When $F=C$, we get a similar conclusion. When $E=A$, we get $F=K=D=A$. Thus $X=AH\cap\odot(ABC)$ and the result is obvious. When $BCEF$ is concyclic, we get $D=AH\cap\odot(ABC)$ therefore $X=A$. The result is obvious as $K$ lie on $A$-altitude. When $EF\parallel BC$, point $D$ becomes $A$-antipode and $X$ is located so that $AXBC$ is isosceles trapezoid. Thus if $M,N$ be the midpoints of $BC,EF$, then it suffices to show that $$\frac{AH}{AA'} = \frac{KN}{HN}$$but this is obvious as $KN : HN = OM : H_aM = AH : AA'$. ($H_a$ is foot from $A$-altitude).

When we have $D_A\in EF$ at #10, we can completes the proof from https://artofproblemsolving.com/community/c6h1429293p8050432

Solution. Redefine $D$ as $\overline{AK}\cap \Gamma,\ D\neq A$ and let $P$ and $Q$ the intersection points of $HK$ with $\Gamma$, so that $Q$ lies on the shorter arc $BC$. Clearly $$\angle KFH=\angle HEK=90^\circ-\angle A=\angle FCH=\angle HBE$$so $KF$ is tangent to $(CFH)$ and $KE$ is tangent to $(BEH)$. Since $KE^2=KF^2$, we conclude that $K$ lies on the radical axis of these two circles. Define $Q'$ as its second common point. We get $$\angle CQ'B=\angle HQ'B+\angle CQ'H=\angle HEA+\angle AFH=180^\circ-\angle A$$so $Q'$ lies on $\Gamma$, thus implying that $Q=Q'$. Then $$KA^2=KE^2=KH\cdot KQ$$hence, $\angle QPD=\angle QAD=\angle KHA$, so $PD\parallel AH$ and thus $PD\perp BC$, so our definition of $D$ coincides with the original one. $\blacksquare$

Here's my solution: WLOG assume that $AB<AC$ and $AE>AF$ (all other cases can be similarly handled). Let $I$ be the antipode of $A$ in $\odot (AEF),$ and suppose $CH$ meets $\Gamma$ again at $X (\neq C).$ Then $EI \parallel HX$ and $EH=EX$ gives that $$\angle EXH=\angle EHX=\angle IEH=\angle IAF=\angle DAC=\angle DXH \Rightarrow E \in DX$$Defining $Y=BH \cap \Gamma,$ we similarly get that $F \in DY.$ Then $$\angle EDA=\angle XCA=90^{\circ}-\angle BAC=\angle YBA=\angle FDA \Rightarrow DA \text{ bisects } \angle EDF$$Using Fact 5, this gives that $I$ is the incenter of $\triangle DEF$, while $A$ is its $D$-excenter. Let $T$ be the point on $EF$ such that $DT \perp BC.$ Also let $DT \cap \Gamma=P.$ Then $$180^{\circ}-\angle PAH=\angle APD=\angle AYD=\angle AYF=\angle AHF \Rightarrow AP \parallel EF$$This means that $APTH$ is a parallelogram, or equivalently that $PH$ bisects segment $AT.$ So for showing that $K \in PH$ (i.e. $PH$ bisects $AI$), we just need to prove that $KH \parallel IT.$ Taking $AD \cap EF=Z$, and since $\angle HAZ=\angle TDZ$, we have $$\triangle HAZ \sim \triangle TDZ \Rightarrow \frac{HA}{TD}=\frac{AZ}{DZ}=\frac{KI}{DI}=\frac{KA}{DI}$$where the second equality is well known (wrt $\triangle DEF$). Again from $\angle HAK=\angle TDI$, we get $$\triangle HAK \sim \triangle TDI \Rightarrow \angle AKH=\angle DIT \Rightarrow KH \parallel IT \quad \blacksquare$$

I haven't seen a solution that utilizes the nice Miquel point in the problem yet, so here's one that does. [asy][asy] size(10cm); defaultpen(fontsize(10pt)); pen ct=chartreuse; pen gr=green; pen cy=cyan; pen sg = springgreen; pair A, B, C, H, D, E, F, K, T, N, S, h, P, X, Q; A = dir(110); B = dir(210); C = dir(330); H = orthocenter(A, B, C); X = dir(10); E = extension(H, X, A, B); F = extension(H, X, A, C); K = circumcenter(A, E, F); h = 2*(foot(H, B, C)) - H; D = IP(Line(A, K, 20), circumcircle(A, B, C), 1); N = extension(A, K, E, F); S = IP(circumcircle(A, B, C), circumcircle(A, E, F), 1); P = extension(E, F, B, C); T = extension(P, S, D, foot(D, B, C)); Q = IP(Line(T, K, 20), circumcircle(A, B, C), 1); draw(A--B--C--A, ct); draw(circumcircle(A, B, C), sg); draw(P--F, ct); draw(A--D, gr); draw(P--T, gr); draw(P--D, gr); draw(circumcircle(A, E, F), sg); draw(circumcircle(A, H, N), sg); draw(Q--T, ct); draw(D--T, ct); draw(P--B, ct); draw(A--T, ct); draw(A--h, ct); draw(circumcircle(S, H, h), dashed+cy); draw(circumcircle(E, B, S), dashed + cy); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$K$", K, dir(K)); dot("$P$", P, dir(P)); dot("$D$", D, dir(D)); dot("$T$", T, dir(T)); dot("$N$", N, dir(250)); dot("$Q$", Q, dir(Q)); dot("$H'$", h, dir(h)); dot("$S$", S, dir(172)); dot("$H$", H, dir(105)); [/asy][/asy] Let $T$ be the point on $\odot (ABC)$ such that $AT \parallel EF$ and let lines $EF$ and $BC$ meet at $P$. Line $TP$ intersects $\odot (ABC)$ again at $S$, lines $AK$ and $EF$ meet at $N$, and let $H'$ be the reflection of $H$ about $BC$. Claim 1: $DH'$ and $EF$ are reflections of each other about $BC$. Let $H_B$ and $H_C$ be the reflections of $H$ about $AC$ and $AB$ respectively. Then $H_BF$ and $H_CE$ meet at a point $D'$ satisfying $$2\measuredangle BAC = \measuredangle H_BDH_C = \measuredangle FDE,$$so $D'$ lies on $\odot (ABC)$ as well as $\odot (EKF)$. Then since $$\measuredangle H_BDH_C = 2\measuredangle H_BDA = 2\measuredangle FDK$$it follows that $D'$ lies on $AK$ and thus $D = D'$. Then $D$ is just the Anti-Steiner point of line $EF$, so $EF$ and $DH'$ are reflections in line $BC$. $\blacksquare$ Claim 2: S is the Miquel point of both $HH'DN$ and $BEFC$. By Claim 1, it suffices to check that $S$ lies on both $\odot (PHH')$ and $\odot (PEB)$. For the former, by construction, we have $$\measuredangle HPS = \measuredangle ATS = \measuredangle AH'S$$so $S$ lies on $\odot (PHH')$. For the latter, simply note that $$\measuredangle EPA = \measuredangle HPA = \measuredangle HH'S = \measuredangle ABS$$and we're done. $\blacksquare$ Claim 3: $\odot (AHN)$ and $\odot (ABC)$ are inverses about $\odot (AEF)$. First, note that Claim 2 implies that $S$ lies on both $\odot (AEF)$ and $\odot (AHN)$. Let $\infty_{HH_C}$ be the point at infinity along $HH_C$; then $$-1 = (H, H_C; HH_C \, \cap \, AB , \infty_{HH_C}) \overset{E}{=} (N, D; A, AK \, \cap \, \odot(AEF))$$so by a well-known property of harmonic divisions it follows that $N$ and $D$ are inverses about $\odot (AEF)$. Finally, note that $A$ and $S$ are fixed upon inversion about $\odot(AEF)$, so the result easily follows. Let the inverse of $H$ about $\odot(AEF)$ be $Q \in \odot(ABC)$. $\blacksquare$. To finish up, let $T'$ be the second intersection of line $QH$ with $\odot(ABC)$. Then $HN$ and $QD$ are antiparallel in $\angle HKN$ by inversion, as are $QD$ and $AT'$, so it follows that $T' = T$. Finally, since the reflection of $H'D$ about $BC$ is parallel to $AT$, it follows that $ATDH'$ is an isosceles trapezoid and thus $TD \perp BC$. So the desired point of intersection is precisely $T$. Remark: Interestingly enough, the first three claims alone are enough to instantly solve 2005 ISL G4.

And yes, complex numbers works too... First if $\overline{AD} \perp \overline{BC}$ there is nothing to prove, so we assume this is not the case. %Also, if $D = B$ then $D = E = B$ and $F$ is the foot of the $D$-altitude, %meaning $K$ is the midpoint of $\overline{AB}$, %and so the concurrency point is the antipode of $C$. %Thus we will assume none of these are 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 \[ \triangle AKH \sim \triangle WOG \qquad(\heartsuit) \]for some point $G$ lying on line $\overline{BC}$. [asy][asy] pair A = dir(130); pair B = dir(210); pair C = dir(330); pair D = dir(265); pair H = orthocenter(A, B, C); pair X = -B*C/D; pair K = extension(A, D, H, X); pair O = origin; pair E = extension(A, B, H, foot(H, A, -X)); pair F = extension(E, H, A, C); draw(A--B--C--cycle, orange); filldraw(unitcircle, invisible, orange); draw(A--D, red); draw(D--X, red); draw(E--F, orange); pair S = (B*F-E*C)/(B+F-E-C); filldraw(circumcircle(A, E, F), invisible, lightblue); draw(H--X, lightblue); pair W = -D; pair G = S+(H-S)*(C-S)/(F-S); filldraw(S--A--K--H--cycle, invisible, blue+0.8); filldraw(S--W--O--G--cycle, invisible, red+0.8); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$H$", H, dir(270)); dot("$X$", X, dir(X)); dot("$K$", K, dir(170)); dot("$O$", O, dir(315)); dot("$E$", E, dir(E)); dot("$F$", F, dir(F)); dot("$S$", S, dir(S)); dot("$W$", W, dir(W)); dot("$G$", G, dir(G)); /* TSQ Source: A = dir 130 B = dir 210 C = dir 330 D = dir 265 H = orthocenter A B C R270 X = -B*C/D K = extension A D H X R170 O = origin R315 E = extension A B H foot H A -X F = extension E H A C A--B--C--cycle orange unitcircle 0.1 yellow / orange A--D red D--X red E--F orange S = (B*F-E*C)/(B+F-E-C) circumcircle A E F 0.1 lightcyan / lightblue H--X lightblue W = -D G = S+(H-S)*(C-S)/(F-S) S--A--K--H--cycle 0.1 blue / blue+0.8 S--W--O--G--cycle 0.1 orange / red+0.8 */ [/asy][/asy] 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{align*} k + ad \overline k &= a+d \\ \frac{k+\frac{bc}{d}}{a+b+c+\frac{bc}{d}} &= \frac{\overline k + \frac{d}{bc}}{\frac1a+\frac1b+\frac1c+\frac{d}{bc}} \\ \iff \frac{\frac1a+\frac1b+\frac1c+\frac{d}{bc}}{a+b+c+\frac{bc}{d}} \left( k + \frac{bc}{d} \right) &= \overline k + \frac{d}{bc} \end{align*}Adding $ad$ times the last line to the first line and cancelling $ad \overline k$ now gives \[ \left( ad \cdot \frac{\frac1a+\frac1b+\frac1c+\frac{d}{bc}} {a+b+c+\frac{bc}{d}} + 1 \right) k = a + d + \frac{ad^2}{bc} - abc \cdot \frac{\frac1a+\frac1b+\frac1c+\frac{d}{bc}}{a+b+c+\frac{bc}{d}} \]or \begin{align*} \left( ad \left( \frac1a+\frac1b+\frac1c+\frac{d}{bc} \right) + a+b+c+\frac{bc}{d} \right) k &= \left( a+b+c+\frac{bc}{d} \right) \left( a + d + \frac{ad^2}{bc} \right) \\ &- abc \cdot \left( \frac1a+\frac1b+\frac1c+\frac{d}{bc} \right). \end{align*}We begin by simplifying the coefficient of $k$: \begin{align*} ad \left( \frac1a+\frac1b+\frac1c+\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} \left[ bc + d(b+c+d) \right] \\ &= \frac{(ad+bc)(d+b)(d+c)}{bcd}. \end{align*}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( \frac1a+\frac1b+\frac1c+\frac{d}{bc} \right) \\ &= \left( a^2+ab+ac+\frac{abc}{d} \right) + \left( da+db+dc+bc \right) \\ &\quad+ \left( \frac{a^2d^2}{bc} + \frac{ad^2}{c} + \frac{ad^2}{b} + ad \right) - \left( ab+bc+ca+ad \right) \\ &= 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 \overline g &= \frac{bc \cdot \frac{ac+ab+ad+bc}{abcd}}{\frac{a-d}{ad}} = -\frac{ab+ac+ad+bc}{d-a}. \\ \implies g + bc \overline g &= \frac{(d-a)(b+c)}{d-a} = b+c. \end{align*}Hence $G$ lies on $BC$ and this completes the proof.

MarkBcc168 wrote: [*] This means that point $K$ is induced by two pencils from $\infty_{\perp AB}$ and $\infty_{\perp AC}$ i.e. $K$ moves along a hyperbola $\mathcal{H}$. As $AD$ has degree $1$, $K$ has degree $2$. Sorry but how do 2 pencils induce a hyperbola, also where can I learn more about hyperbolas including why does taking the isogonal conjugate of a line make a hyperbola?

Quote: When $BCEF$ is concyclic, we get $D=AH\cap\odot(ABC)$ therefore $X=A$. The result is obvious as $K$ lie on $A$-altitude. When $EF\parallel BC$, point $D$ becomes $A$-antipode and $X$ is located so that $AXBC$ is isosceles trapezoid. Thus if $M,N$ be the midpoints of $BC,EF$, then it suffices to show that $$\frac{AH}{AA'} = \frac{KN}{HN}$$but this is obvious as $KN : HN = OM : H_aM = AH : AA'$. ($H_a$ is foot from $A$-altitude). Nice solution, but if $AB=AC$, these two cases coincide and you should consider one more case.

Quote: Nice solution, but if $AB=AC$, these two cases coincide and you should consider one more case. The $AB=AC$ case follows from continuity.

Let $\overline{HK}$ intersect $(ABC)$ at $P$ and $Q$, with $P$ on the same side of $\overline{BC}$ as $A$. Claim. Define $Q' = (EHB) \cap (FHC)$. Then $Q' = Q$. Proof. Notice that $\angle BQ'H + \angle CQ'H = 180^\circ -\angle A$, hence $Q'$ lies on $(ABC)$. Now, $\overline{Q'H}$ bisects $\angle EQ'F$, so it passes through the arc midpoint $K'$. Because $\angle EK'F = 180^\circ - \angle EQF = 180^\circ - \angle A$, follows that $K' = K$ and thus $Q' = Q$. $\blacksquare$ Now, notice that $\overline{KE}$ is tangent to $(EHQ)$ as $\angle KEF = \angle HQE$. Thus $KH \cdot KQ = KE^2 = KA^2$, implying $\angle QPD = \angle QAD = \angle QHP$, i.e. $\overline{AH} \parallel \overline{DP}$. The result then follows. Remark: When solving this problem a few days ago, some part of me vaguely remembers proving that $EQDFK$ is cyclic, but this fact isn't even necessary!?

Let $HK$ meet $\Gamma$ at $P, Q$, with $P, D$ on opposite sides of $BC$. We can see that $\angle KEF = 90 - \angle A = \angle EBH$ gives us that $KE$ is tangent to $(EBH)$. Similarly, $KF$ is tangent to $(FCH)$, and since $KE^2 = KF^2$, we have that $KH$ is the radical axis of $(EBH)$ and $(CFH)$. Now, let $Q' = (EBH) \cap (CFH)$. Clearly we have $\angle BQ'C = \angle BQ'H + \angle HQ'C = 180 - (180 - \angle AEF - \angle AFE) = 180 - \angle A$, so $Q' \in (ABC)$. Thus $Q'H$ is the radical axis of $(EBH), (CFH)$, so $Q' \in HK$, so $Q' = Q$. Then, we have \[\angle(PD, BC) = \angle DAC + \angle BQP = \angle DAC + \angle AEF = \angle KEF + \angle EAF = 90 \] and we are done.

Define $M$ as the intersection of $(ABC)$ as the perpendicular from $D$ to $BC$, and $N = MH \cap (ABC)$. $BNHE$ and $CNHF$ are cyclic, as $\angle AEF = 90 - \angle CAD = \angle MCB = \angle HNB$. $KE$, $KF$ are tangent to $BNHE$, $CNHF$ as $\angle HBE = 90 - \angle FAE = \angle HEK$. Thus $K$ lies on the radical axis of the two circles, which is $NHM$. $\blacksquare$

Let $O$ be the circumcenter of $(ABC).$ Denote by $\ell$ the line through $D$ perpendicular to $\overline{BC}.$ Fix $\triangle ABC$ and animate $E$ projectively on $\overline{AB}.$ Then $H$ is fixed and $F,K,D,E$ move as a function of $E.$ Consider the line through the midpoint of $\overline{AE}$ and the point at infinity in the direction perpendicular to $\overline{AB},$ and similarly the line through the midpoint of $\overline{AF}$ and the point at infinity in the direction perpendicular to $\overline{AC}.$ These lines each have degree $1,$ so $\deg(K)=2.$ Consequently we have $\deg(D)=2,$ and note that the map $D \mapsto G_1$ is identical to taking a reflection about the line through $O$ parallel to $\overline{BC},$ so we know $\deg(G_1)=2.$ On the other hand, consider the degree of $G=\overline{HK} \cap \ell.$ Since $\deg(K) = 2$ and $\deg(D)=2,$ it follows $\deg(G)=4.$ Thus, it suffices to verify $G \equiv G_1$ for five cases of $E.$ But this is not hard to do. The following five points are easy to verify: $E=A$ obvious $E=B$ is obvious $E$ the point such that $\overline{EH} \parallel \overline{AC}.$ $E=\text{foot}(C,AB)$ $E$ the point such that $\overline{EH}$ is antiparallel to $\overline{BC}.$ This completes the proof.

Let $R$ be the image of $H$ under inversion about $(AEF)$. We claim that $R \in \Gamma$. Indeed, \[ \measuredangle EQK = \measuredangle KEH = 90 - \measuredangle FEA = \measuredangle EBH \]so $(BEHR)$, and similarly $(CFHR)$; by the Miquel config in $\Delta AEF$ we obtain $R \in \Gamma$. Now if $RHK$ meets $\Gamma$ again at $T$, then \[ \measuredangle HAK = \measuredangle KQA = \measuredangle TDA \]so $AH \parallel TD$, so $TD \perp BC$ and $T$ is the desired point.

Denote $\angle HAE = \alpha$, $\angle HAK = \beta$ and $\angle KEF = \angle KFE = \gamma$ $\Rightarrow$ $\angle KAE = \angle AEK = \alpha + \beta$ and $\angle KAF = \angle KFA = 90 - \alpha - \beta - \gamma$. Now $\angle HBE = 180 - 90 - \angle BAC = 90 - (90 - \gamma) = \gamma$ $\Rightarrow$ $\angle KEF = \angle EBH = \gamma$ $\Rightarrow$ KE is tangent to (BEH). Similarly $\angle HCF = 180 - 90 - \angle BAC = 90 - (90 - \gamma) = \gamma$ $\Rightarrow$ $\angle KFE = \angle FCH = \gamma$ $\Rightarrow$ KF is tangent to (CFH). Now let $(BEH) = \omega_1$ and $(CFH) = \omega_2$. PoP of K w.r.t $\omega_1$ is equal to $KE^2$. PoP of K w.r.t $\omega_2$ is equal to $KF^2$, but KE = KF $\Rightarrow$ $KE^2 = KF^2$ $\Rightarrow$ $K \in \rho (\omega_1, \omega_2)$ $\Rightarrow$ $H \in \omega_1$, $H \in \omega_2$ $\Rightarrow$ $H \in \rho (\omega_1, \omega_2)$ $\Rightarrow$ $KH \equiv \rho (\omega_1, \omega_2)$. Denote the second point of intersection of $\omega_1$ and $\omega_2$ be Y. Now $\angle BYH = \angle AEH = \alpha + \beta + \gamma$ and $\angle HYC = \angle HFA = 90 - \alpha - \beta$ $\Rightarrow$ $\angle BYC = \angle BYH + \angle HYC = \alpha + \beta + \gamma + 90 - \alpha - \beta = 90 + \gamma$ $\Rightarrow$ $\angle BAC + \angle BYC = 90 - \gamma + 90 + \gamma = 180^{\circ}$ $\Rightarrow$ BACY is cyclic $\Rightarrow$ $Y \in \Gamma$. We know that $HY \equiv \rho (\omega_1, \omega_2) \equiv KH$ $\Rightarrow$ K, H, Y lie on one line. Let $HK \cap \Gamma = P$. Now it is left to show that $PD \perp BC$. Let $PD \cap BC = Z$ $\Rightarrow$ we now want to show that $\angle PZB = 90^{\circ}$. We have that $\angle PZB = 180 - \angle BPZ - \angle ZBP = 180 - \angle BPD - \angle CBP = 180 - \angle BAD - \angle CYP = 180 - (\alpha + \beta) - \angle HFA = 180 - (\alpha + \beta) - (90 - \alpha - \beta) = 180 - \alpha - \beta - 90 + \alpha + \beta = 90^{\circ}$ $\Rightarrow$ $\angle PZB = 90^{\circ}$ which is exactly what we wanted to prove. We are ready.

Move $E$ along $\overline{AB}$ with degree $1.$ Then $E \mapsto{H} F$ by perspectivity. Note that the midpoint of of $AE$ and $AF$ move with degree $1,$ and projecting from a point at infinity for both of them, the moving line that is the perpendicular bisector of $AE,AF$ has degree $1+1-1=1$ by Zack's lemma. Thus, $K$ moves with $1+1-0=2.$ Let $X$ be where the intersection of the line through $D$ that is perpendicular to $\overline{BC}$ intersects $(ABC).$ Note that $K\mapsto{A} D \mapsto{\infty_{\overline{AH}}} X.$ Thus, $X$ moves with degree $2$ as well. Therefore, as $\text{deg} H+ \text{deg}K+\text{deg}X=4,$ we simply need to check $4+1=5$ cases. These are $E=A.$ $F=C.$ $EF \| BC$ because $\triangle AKH \sim \triangle AOH_A \sim \triangle DKG.$ $EFBC$ concyclic(Isogonal conjugate of orthocenter is circumcenter). $E=B$(Reflection of orthocenter).$\blacksquare$

Let $DO \cap \Gamma = D'$, so now we want to show that D', H, K lie on one line. Since $AH \parallel DD'$ and $\angle HAK = \angle KDD'$, to prove that D', H, K lie on one line it is enough to show that $\triangle AKH \sim \triangle DKD'$ $\Leftrightarrow$ $\frac{AK}{DK} = \frac{AH}{DD'}$. Let $\angle EAH = \alpha$, $\angle HAK = \beta$, $\angle FEK = \gamma$. Now we know that $AH = 2R\cos \angle BAC = 2R\cos (90 - \gamma) = 2R\sin \gamma$. From $\triangle DD'C$, we have that $DD' = 2R\sin \angle D'CD = 2R\sin (2\alpha + 2\beta + \gamma)$ $\Rightarrow$ $\frac{AH}{DD'} = \frac{2R\sin \gamma}{2R\sin (2\alpha + 2\beta + \gamma)} = \frac{\sin \gamma}{\sin (2\alpha + 2\beta + \gamma)}$. By law of sines on $\triangle AEH$, we get $\frac{AE}{\sin \angle AHE} = \frac{AH}{\sin \angle AEH}$ $\Rightarrow$ $\frac{AE}{\sin (180 - 2\alpha - \beta - \gamma)} = \frac{AH}{\sin (\alpha + \beta + \gamma)}$ $\Rightarrow$ $AE = \frac{AH \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma)} = \frac{2R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma)}$. Let P be the midpoint of AE $\Rightarrow$ $AP = \frac{R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma)}$. For $\triangle APK$, we have that $\frac{AP}{AK} = \cos \angle PAK = \cos (\alpha + \beta)$ $\Rightarrow$ $AK = \frac{AP}{\cos (\alpha + \beta)}$ $\Rightarrow$ $AK = \frac{R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta)}$. Now from $\triangle ADC$, we have that $AD = 2R\sin \angle ACD = 2R \sin (2\alpha + \beta + \gamma)$ $\Rightarrow$ $DK = AD - AK = \frac{2R\sin (2\alpha + \beta + \gamma) \cdot \sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta)} = \frac{R\sin (2\alpha + \beta + \gamma) \cdot (2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma)}{\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta)}$ $\Rightarrow$ $\frac{AK}{DK} = \frac{R\sin \gamma \cdot \sin (2\alpha + \beta + \gamma)}{R\sin (2\alpha + \beta + \gamma) \cdot (2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma)} = \frac{\sin \gamma}{2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma}$. We want to show that $\frac{AK}{DK} = \frac{AH}{DD'}$ $\Leftrightarrow$ $\frac{\sin \gamma}{2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma} = \frac{\sin \gamma}{\sin (2\alpha + 2\beta + \gamma)}$ $\Leftrightarrow$ $2\sin (\alpha + \beta + \gamma) \cdot \cos (\alpha + \beta) - \sin \gamma = \sin (2\alpha + 2\beta + \gamma)$, which is obvious by trig formula $\Rightarrow$ $\frac{AK}{DK} = \frac{AH}{DD'}$, $\triangle AKH \sim \triangle DKD'$ $\Rightarrow$ D', H and K lie on one line and we are ready.

I also like moving points.

Cool Problem! My first good problem of TSTST I did a few months ago Solution: Let $L$ be such that $DL\perp BC$ and $LH \cap \odot(ABC)=J$. So we need to show that $(HEB)\cap(HEC)=J$ and $K\in LHJ$. We angle chase to get $$\measuredangle CJH =\measuredangle CDL=90^{\circ}- \measuredangle DCB= 90^{\circ}-\measuredangle KAE= \measuredangle AFE=\measuredangle HFC\implies \odot(CFHJ)$$Similarly we get $ \odot(BEHJ)$. Now to prove $K\in LHJ$, note that this can be done if $KE$ and $KF$ are tangents this is because by Radical axis and Power of Point we will have $\text{Pow}_{\odot(BEHJ)}(K)=\text{Pow}_{\odot(CFHJ)}(K)$ and we get $K\in LHJ$, so by angle chasing we have $\measuredangle KEH = 90 - \measuredangle BAC = \measuredangle EBH$ and we are done!

Really nice problem, although trying to not get circular arguments was kinda tricky. Let $D' \in (ABC)$ be the Anti-Steiner point of $\overline{EF}$ w.r.t $(ABC)$, $G = (AEF) \cap \overline{AK}$ and $P$ be the intersection of $(ABC)$ and the line through $D$ perpendicular to $\overline{BC}$. Notice that the reflection of $\overline{EF}$ over $\overline{AB}, \overline{AC}$ passes through $D'$, so they are the external bisectors of $\angle D'EF$ and $\angle D'FE$ respectively, thus $A$ is the $D'$-excenter of $\triangle D'EF$. Then, we have $G$ is the incenter of $\triangle D'EF$, and since $EK = FK$, we have $\overline{D', G, K}$. Hence, we have $\overline{A, K, G, D'}$, so $D = D'$. Let $Q = \overline{HK} \cap (DEF)$. Then $$\angle EGH = \angle EFK = 90 - \angle A = \angle EBH$$so $BQHE$ is cyclic. Similarly, $CQHF$ is cyclic. After that, because $$\angle BQC = \angle BQH + \angle CQH = \angle AEF + \angle AFE = 180 - \angle A$$so $Q \in (ABC)$. Let $H'$ denote the reflection of $H$ across $BC$, which is also the intersection of $\overline{AH}$ and $(ABC)$. Since $D$ is the Anti-Steiner point of $\triangle ABC$, the reflection of $\overline{EF}$ across $\overline{BC}$ is $\overline{H'D}$. Note that $\overline{AH'} // \overline{DP}$, so $$\angle APD = \angle H'DP = \angle(\overline{EF}, \overline{DP})$$which means that $\overline{AP} // \overline{EF}$. From which we get $$\angle BQH = \angle AEF = 180 - \angle PAE$$Let $\overline{QHK} \cap (ABC) = P'$. This means that $\angle PAE = 180 - \angle BQH = \angle P'AE$, so $P = P'$, i.e. $\overline{Q, H, K, P}$. Therefore, $\overline{H, K, P}$, as desired.

Let the perpendicular from $D$ to $BC$ meet $\Gamma$ at $S$ and let $SH$ meet $\Gamma$ again at $L$. We prove that $K$ lies on $\overline{S-H-L}$. Now note that by some angle chasing, we get that $BEHL$, $CFHL$ are cyclic and that $KE$ and $KF$ respectively are tangent to these circles so $K$ must lie on the radical axis which is $LH$ and thus we are done.

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Define point $T$ on $(ABC)$ such that: $DT \perp BC$. Let $TH$ intersect $(ABC)$ at $S$. Claim: $BEHS$, $CFHS$ are cyclic. Proof: Notice that: $$\angle HSB = \angle TDB = 90^\circ - \angle DBC = \angle 90^\circ - \angle DAC = 90^\circ - \angle KAF = \angle AEH$$which implies $EHSB$ is cyclic. Similarly, $CFHS$ is cyclic and we are done. Claim: $KE, KF$ are tangent to $(BEH), (CFH)$ respectively. Proof: Notice that: $KE=KF$ and $\angle KEH = \angle HBE = 90^\circ - A$ which concludes the $KE$ to be tangent to $(BEH)$. Similarly, $KF$ is tangent to $(CFH)$ and the claim is proved. Note that: $HS$ is the radical axis of $(BEH), (CFH)$ and $T, K \in HK$ (due to the above two claims). We are done.

gives us that lines $EF$ and $H'D$ are reflections across $\overline{BC}$. So, if $X$ is the intersection of lines $EF$ and $DD'$, it follows that $XHH'D$ is an isosceles trapezoid. Moreover, $AD'XH$ is a parallelogram. Claim: $D$ lies on the circumcircle of $EKF$. Proof: Firstly, $BEXD$ is cyclic since \[\angle EBD + \angle DXE = \angle AH'D + \angle HXD = 180^{\circ}.\]So, we have \begin{align*}\angle FED &= \angle XED = \angle XBD = 2 \angle CBD \\ &= 2 \angle CAK = 180^{\circ} - \angle AKF = \angle FKD. ~ \blacksquare \end{align*} Let $I$ be the incenter of $\triangle DEF$, which coincides with the second intersection of line $AD$ with $(AEF)$. Let $G$ be the intersection of line $AD$ with line $EF$. Since $A$ is the $D$-excenter of $\triangle DEF$, we have by $\sqrt{ef}$ inversion that \begin{align*} DA \cdot DI = DG \cdot DK & \implies \tfrac{DK}{IK} = \tfrac{DA}{GA} \\ & \implies \tfrac{DK}{AK} = \tfrac{DG + GA}{GA} \\ & \qquad \qquad ~ ~ = \tfrac{DX + AH}{AH} \\ & \qquad \qquad ~ ~ = \tfrac{DD'}{AH},\end{align*}which is sufficient to imply the collinearity.