Let us invert trough $H$ with power $-AH\cdot HD$, so that $A,B,C\mapsto D,E,F$ respectively, and $P$ must be the antipodal point of $H$ on $(HBC)$. It will suffice to prove that $(DT^*L^*)$ and line $P^*K$ are tangent at $K$ (note that $P^*K$ is parallel to $AD$, by considering the symmetry with respect to $M_{BC}$, so indeed $P^*K$ is the inverse of $\omega$), so that the the point of tangency trivially lies on $HK$. By the ciclicities, we have $\measuredangle DL^*T^*=\measuredangle DL^*H=\measuredangle DBH=\measuredangle CBH$ and $\measuredangle L^*T^*D=\measuredangle HT^*D=\measuredangle HCD=\measuredangle HCB$. This implies that $HBC$ and $DL^*T^*$ are (negatively) similar. Now, $\frac{DK}{DL^*}=\frac{P^*H\sin \angle KP^*H}{BH\sin \angle DHL^*}=\frac{P^*H}{BH}$ (the first equality is by law of sines and by the fact that if $Z=P^*H\cap BC$ then $DK=DZ+ZK=HZ\sin \angle DHZ+P^*Z\sin\angle ZP^*K=(HZ+ZP^*)\sin\angle KP^*H$), and so we also have that $BHP^*$ and $L^*DK$ are (negatively) similar (note they also have the angle between the similar sides in common). Therefore, uniting the two facts, the two quadrilaterals $HBP^*C$ and $DL^*KT^*$ are (negatively) similar. Thus, since $P^*$ and $H$ are antipodals, it follows that $D$ and $K$ are also antipodals, and so their tangents are parallel. Also $\measuredangle L^*T^*D=\measuredangle HCB=\frac{\pi}{2}-\measuredangle BCP^*=\frac{\pi}{2}-\measuredangle BHP^*=\frac{\pi}{2}-\measuredangle BHL^*=\measuredangle L^*BH=\measuredangle L^*DH$, which implies $(L^*DT^*)$ is tangent to $AH$. Finally this implies that $(L^*DT^*)$ is tangent to $P^*K$ at $K$, and thus $\omega$ is tangent to $(ATL)$ at $K^*$, which clearly lies on $KH$.

Let $O$ be the center of $\Omega \equiv \odot(ALT)$ ; $\Gamma \equiv \odot(AH)$ ; $M$ be the common midpoint of segments $BC,DK$. Note that $\angle ALT = \angle HAT = 90^\circ - \angle B$, so $\overline{AH}$ is tangent to $\Omega$ at $A$. This gives that $\overline{AO} \parallel \overline{BC}$. Let $Y \in \Omega$ such that $\overline{HY},\overline{HA}$ are tangent to $\Omega$ ; $X = \overline{AY} \cap \overline{OH}$ (which lies on $\Gamma$) ; $Z$ be antipode of $A$ wrt $\Omega$ ; $Q = \Omega \cap \Gamma \ne A$ (which lies on $\overline{HZ}$) ; $R = \overline{XX} \cap \overline{HH}$. We will prove that $\boxed{Q \text{ is the desired tangency point and } Q \in \overline{HK}}$. [asy][asy] size(250); pair B = dir(-150),C=dir(-30),A=dir(105),H=A+B+C,D=foot(A,B,C),E=foot(B,C,A),F=foot(C,A,B),K=B+C-D,Z=extension(K,H,A,dir(75)),M=1/2*(B+C); path Omega = circle(1/2*(A+Z),1/2*abs(A-Z)); pair P = foot(H,E,F),L=extension(H,P,A,C),T=extension(H,P,A,B),O=1/2*(A+Z); path Gamma = circle(1/2*(A+H),1/2*abs(A-H)); pair Q = 2*foot(A,O,1/2*(A+H))-A,X=foot(A,O,H),Y=2*X-A,R=extension(A,Q,E,F); path omega = circle(1/2*(R+H),1/2*abs(R-H)); draw(Omega^^Gamma^^omega,purple); dot("$A$",A,dir(40)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$M$",M,dir(M)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$K$",K,dir(K)); dot("$Z$",Z,dir(Z)); dot("$H$",H,dir(70)); dot("$P$",P,dir(P)); dot("$L$",L,dir(L)); dot("$T$",T,dir(160)); dot("$O$",O,dir(O)); dot("$Q$",Q,dir(70)); dot("$X$",X,dir(150)); dot("$Y$",Y,dir(-150)); dot("$R$",R,dir(R)); draw(A--B--C--L,fuchsia); draw(H--L^^Z--K^^1.2*Y-0.2*H--H^^D--1.2*A-0.2*H,orange); draw(1.2*X-0.2*R--R--1.2*H-0.2*R^^A--R--E,royalblue); draw(Z--1.1*A-0.1*Z^^M--O--1.2*Q-0.2*O,green); draw(A--Y^^B--E^^C--F,brown); label("$\omega$",1/2*(R+B),purple); label("$\Omega$",1.45*foot(O,L,Z)-0.45*O,purple); label("$\Gamma$",1.75*P-0.75*B,purple); [/asy][/asy] Note that $$-1 = (A,Y ; L,T)_{\Omega} \stackrel{A}{=} (H,X;E,F)_{\Gamma} \stackrel{H}{=} (\infty_{\overline{BC}},\overline{HX} \cap \overline{BC} ; B,C)$$So $ \overline{HX} \cap \overline{BC} \equiv M$. Then homothety at $H$ gives points $Z,H,K$ are collinear, in particular $\boxed{Q \in \overline{HK}}$. Note that $\Omega,\Gamma$ are orthogonal so tangents at $A,Q$ to $\Omega$ concur at $O$. thus $(H,X ; A,Q)_{\Gamma} = -1$ which gives $R \in \overline{AQ}$. Also, as $(H,X ; E,F)_{\Gamma} = - 1$, so $R \in \overline{EF}$. Now note that $\odot(RH)$ passes through $P$ and is also tangent to $\overline{AH}$, in particular $\omega \equiv \odot(RH)$. Then it easy to see that a homothety at $Q$ sends $\Omega$ to $\omega$. This completes the proof of the problem. $\blacksquare$ Remark: By considering homothety at $H$ sending line $BC$ to line $AO$, our whole can be restated in terms on $\triangle ALT$ as follows: Let $\Omega$ be the circumcircle of a $\triangle ALT$ ; $O$ be center of $\Omega$ ; $H = \overline{AA} \cap \overline{LT}$. Define $E,F,P$ as the projections of $H$ onto $\overline{AL},\overline{AT},\overline{EF}$, respecitvely. Let $\omega$ be the circle passing through $H,P$ and tangent to $\overline{AH}$. Prove that $HO$ is the $H$-symmedian of $\triangle HEF$. $\omega$ and $\Omega$ are tangent to each other at some point $Q$ and moreover, $\overline{HQ}$ passes through the antipode of $A$ wrt $\Omega$.

Sinus Law

Let $\omega$ and $(AFE)$ intersect at $Q$, and the line through $H$ parallel to $BC$ cut $\omega$ at $X$. then $A,Q,X$ and $P,F,X$ are collinear. $HF \cap \omega = M$. Then Pascal on $QMHHPX$ shows that $QM \cap HP$ lies on $AB$, which means that $Q,M,T$ collinear. $$\angle TQA = \angle MQX = \angle MHX=\angle FEH = 90- \angle B$$so $Q$ lies on $(ALT)$. Let the midpoint of $AB$ be $W$, $EF \cap BC=O'$, $AX\cap BC = O$. Then by Menelaus' on $\triangle ODA$ wrt line $O'XV$, $$\frac{O'O}{O'D}\cdot \frac{DV}{VA}\cdot \frac{AX}{XO}=1$$Which means, $$\frac{O'O}{O'D}=\frac{AV}{VD}\cdot \frac{OX}{XA}= \frac{AV}{VD}\cdot \frac{DH}{HA}$$where $V=AH\cap EF$. To prove that $K$ lies on $QH$, it suffices to prove that $DB\cdot DC= DO \cdot DK$.On the other hand, $$\frac{AV}{VD} \cdot \frac{DH}{HA} = \frac{AF}{FD}\cdot \frac{\sin C}{\sin 2C} \cdot \frac{FD}{FA} \cdot\frac {\cos C}{\sin 90} = \frac{1}{2} \Rightarrow \frac{O'O}{O'D}=\frac{1}{2}$$Then $$DO\cdot DK = \frac{1}{2}DO' \cdot 2\cdot DW= DB\cdot DC$$which concludes our proof.

My solution at the contest and without law of sines. Let $M$ be the intersection of $\omega$ with the circle of diameter $AH$. We will prove that $M$ is the requerid tangency point. We see that the points $E$ and $F$ lies on the circle of diameter $AH$ $(\angle HFA = \angle HEA = 90^{\circ} )$ Claim: Let $A’$ be point diametrically opposite to $A$ on the circumcircle of $\triangle ABC$ and let $U$ be midpoint de $BC$, so $U$, $H$ and $A’$ are collinears. Proof: $BH \perp AC$ and $A’C \perp AC$ $\Rightarrow BH \parallel A’C$. Analogously $CH \parallel A’B$ so $HCA’B$ is a parallelogram whose diagonals intersect at their midpoint, which is $U$. $\square$. Now, let $N$ be the midpoint de $AH$ so $N$ es circumcenter of the $\triangle AFE$, which is circle of diameter $AH$. Since $HA$ is tangent to $\omega$ so $NH$ is tangent too but $NH = NM = NA = NE = NF$ so $NM$ is tangent to $\omega$. Notice that $\angle BAD = \angle TAN = 90^{\circ} - \angle ABC$ but $\angle ALT = \angle ELP = 90^{\circ} - \angle AEF = 90^{\circ} - \angle ABC$ (since $EFBC$ is cyclic). So $\angle ALT = \angle TAN$ therefore $AH$ is tangent to circumcircle of $\triangle ATL$. Claim:The quadrilateral $LEPM$ is cyclic Proof: From circle of diameter $AH$ and $\omega$ we get $\angle LPM = \angle TPM = 180^{\circ} -\angle MPH = \angle MHA = \angle MEA = \angle MEL$ $\square$ As $LEPM$ and $AMFE$ are cyclic, it turns out that $M$ is the Miquel point of complete quadrilateral $ATPE$. Hence the quadrilaterals $TPFM$ and $ATML$ are cyclic, so $M$ lie on the circumcircle of $\triangle ATL$. As $NA= NM$ and $AH$ is tangent to circumcircle of $\triangle ATL$ we get that $NM$ is also tangent to said circle. Therefore $\omega$ and circumcircle of $\triangle ATL$ are tangents to $MN$ at $M$. On the other hand, we will prove that $M$ lie on $KH$. Let $K’ = MH \cap BC$ and line $HU$ meets circumcircle of $\triangle ABC$ for the second time at $X$. Clearly $X$ lie on circle of diameter $AH$. Observe that $\angle HXA = \angle HDU$ and $\angle XHA = \angle DHU$. So $\triangle AXH \sim \triangle UDH \Rightarrow \frac{AX}{DU} = \frac{XH}{DH}$ Also $\angle HUK’ = 180^{\circ} - \angle HUD = 180^{\circ} - \angle HAX = \angle XMH$ and $\angle XHM = \angle K’HU$. So $\triangle XMH \sim \triangle K'UH \Rightarrow \frac{MX}{XH} =\frac{UK’}{HK’}$. We get $ UK’ = \frac{MX \cdot HK’}{XH}$ and $DU = \frac{AX \cdot DH}{XH}$. So we need to prove that $MX \cdot HK’ = AX \cdot DH$. Claim: $AXMH$ is harmonic quadrilateral. Proof:Since the tangent to circle of diameter $AH$ at $H$ is parallel to $BC$, we get $ -1 = (B, C ; M, P_\infty) \stackrel{H}{=} (E, F; X, H) \Rightarrow EHFX$ is a harmonic quadrilateral. So $XX, HH$ and $EF$ are concurrent, say in $I$. Let $R = AH \cap EF$. Clearly $RH^{2} = RP \cdot RI \Rightarrow RH \equiv AH$ is tangent to circumcircle of $\triangle HPI$ so that $\odot (HPI)$ is $\omega$. Thus $\angle HMI = \angle HMA = 90^{\circ} \Rightarrow$ $A, M$ and $I$ are collinears $\square$ Therefore $( A, M; H, X) = -1$ so that $\frac{MX}{AX} = \frac{MH}{AH}= \frac{DH}{HK’}$ (since $\triangle DHK’ \sim \triangle MHA$) Thus $MX \cdot HK’ = AX \cdot DH \Rightarrow UK’ = UD \Rightarrow K’ \equiv K$ and we are done $\blacksquare$

I had to look for the tangency point . Beautiful problem. $HK \cap (AH)=S$. $HK \cap EF=R$. $AS \cap FE=Q$. Note that $QSPH$ is cyclic. $\textbf{Claim.1:}$ $\triangle AST \sim HSF$ and $\triangle ASL \sim HSE$. Proof: $(i) \angle SAT=\angle SHF$ $$(ii) \frac{AS}{AT}=\frac{HS}{HF} \iff \frac{SA}{SH}=\frac{AT}{HF} \iff \frac{DK}{DH}=\frac{AT}{HF} \iff \frac{HF}{HD}=\frac{AT}{DK} \iff \frac{AF}{DC}=\frac{AT}{DK} \iff \frac{DK}{DC}=\frac{AT}{AF} \iff \frac{TF}{AF}=\frac{CK}{CD} \iff \frac{FT}{HF} \cdot \frac{HF}{AF}=\frac{DB}{HD} \cdot \frac{HD}{DC}$$Last equality holds because $HDB \sim HFT$ and $HDC \sim HFA$. $\square$ Because of the Claim.1 $\angle LST =\angle LSA+\angle AST=\angle FSH +\angle HSE=\angle A$. Thus $ALST$ is cyclic. $\textbf{Claim.2:}$ $QH \parallel BC$. Proof: $\frac{QF}{QE}=\frac{SF}{SE}\cdot \frac{AF}{AE}=\frac{RF}{RE}\cdot \frac{HE}{HF} \cdot \frac{AF}{AE}=\frac{HF^2}{HE^2}\cdot \frac{KC}{KB} \cdot \frac{HE}{HF} \cdot \frac{AF}{AE}=\frac{HF^2}{HE^2}\cdot \frac{DB}{DC} \cdot \frac{HE}{HF} \cdot \frac{AF}{AE}=\frac{HF}{HE}\cdot \frac{DB}{DC} \cdot \frac{AF}{AE}=\frac{HF}{HE}\cdot \frac{BF}{EC}=\frac{HF^2}{HE^2}$ $\implies QH$ is tangent to $(AH) \implies QH \parallel BC$.$\square$ Note that $(AH)$ is tangent to $(ALST)$. $\angle SQH +\angle SLA=\angle SAH +\angle SHA=\angle HSA$ as needed.

Proposed by Mahdi Etesamifard - Iran

Let $O\in EF$ such that $OH\perp AH$, let $Q$ be the Miquel point of $BFEC$, $X\in OA$ such that $HX\perp OA$, and let $M$ be the midpoint of $BC$. We have that \[ \angle PXF=\angle PXO-\angle FXO=90+\angle PXH-\angle AEF=90+\angle POH-\angle AEF=90+\angle B-\angle C-\angle B=90-\angle C=\angle FTP \]then $XTPF$ is cyclic, now \[ \angle TXA=\angle 90-\angle HXT=\angle FXH=\angle FAH=\angle TLA \]then $XTLA$ is cyclic and $HA$ is tangent to this circle. Let $\phi$ be the inversion centered at $O$ which fixes $(AEHFXQ)$. Clearly $\phi((OHPX))$ is the line $AH$ and $\phi((TAL))=(TAL)$ then the first result follows. It is a well know result that $QH$ goes through $M$ then $(HF,HE;HQ,HO)$ is harmonic, meaning that $QH$ is the polar of $O$ w.r.t $(AEHFXQ)$, then $(HX,HA;HQ,HO)$ is harmonic which implies that $K,H,X$ are collinear.

Let $(AH)$ intersect $(ATL)$ at $J$.Then,easy to get by angle chase $MJ$ is tangent to $(ATL)$.Also let $AJ$ hit the perp at $H$ to $AH$ at $N$.Then $NJPH$ is cyclic and is $\omega$.Also,$MJ$ is tangent to this circle,so $\omega,(ATL)$ are tangent at $J$.Then let $HJ$ hit $BC$ and the the parallel at $A$ to $BC$ at $K',O$ respectively.Let $Q$ be the midpoint of $DK'$.We have that $AHO,DHK'$ are homothetic,so $HQ$ hits $AO$ at the midpoint say $R$.Well known that $RA$ is tangent to $(AH)$,and so thus is $RJ$.By La Hire,$R$ must lie on the polar of $N$ and thus if $RH \cap (AH)=S$,then $NH,NS$ are tangents.Now,$H,S,E,F$ are harmonic,so tangents at $E$ and $F$ intersect on $RH$.However ,this is known to be the midpoint of $BC$,so $Q$ is the midpoint of $BC$ and $BD=K'C$,so $K=K'$.Also,@3above's proof is scary af

Let $AEHF$ meet $\omega$ at $S$. Let $Q$ be antipode of $H$ in $\omega$. Claim $: A,S,Q$ and $E,P,Q$ are collinear. Proof $:$ Note that $\angle QPH = \angle 90 = \angle HPE$ and $\angle QSH = \angle 90 = \angle ASH$. Claim $: TSFP$ is cyclic. Proof $:$ Note that $\angle SFT = \angle SFA = \angle SHA = \angle SQH = \angle SPT$. Claim $: ATSL$ is cyclic. Proof $:$ Note that $\angle LAS = \angle SFE = \angle SFP = \angle STL$. Claim $: ATL$ and $\omega$ are tangent at $S$. Proof $: \angle TSP = \angle TFP = \angle AFP = \angle FAQ + \angle FQA = \angle TAS + \angle PQS$. Let perpendicular from $A$ meet $EF$ at $R$. Claim $: RHE$ and $FHS$ are similar. Proof $:$ Note that $\angle REH = \angle FEH = \angle FSH$ and $\angle ERH = \angle 180 - \angle QRH = \angle 180 - \angle QAH = \angle 180 - SAH = \angle SFH$. Now Note that $\angle FHS = \angle RHE$. Note that since $H$ is antipode of $A$ in $AEHF$ then $R$ is reflection of $P$ in midpoint of $EF$ so since $EFH$ and $CBH$ are similar we have $\angle EHR = \angle CHK$ so now we have $\angle CHK = \angle SHF$ so $S,H,K$ are collinear.

Let $Q$ be the Miquel point of the quadrilateral $ATPE$. Since the pentagon $AQFHE$ and quadrilateral $ALQT$ are cyclic and $\triangle QTA \sim \triangle QPE$ , we can get : $\angle PQH=\angle EQH-\angle EQP=\angle EAH-\angle ELP=\angle B-\angle C=\angle PHA$ So the circumcircle of triangle $PQH$ is tangent to the line $AH$ and $Q$ lies on the circumcircle of triangle $\triangle ATL$ and $\omega$. Also we know that $\angle ALT=90-\angle B=\angle TAH$ , So $AH$ is common tangent line of the circles $(ATL)$ and $\omega$. Since $\angle AQH=\angle AFH=90$ so if $M$ is the midpoint of $AH$ , then $MQ$ is tangent to both circles and they are tangent at $Q$. Now to prove that $Q , H , K$ are collinear , it's needed to show that $\triangle AQH \sim \triangle KDH$ which is equivalent to : $\frac{DK}{DH}=\frac{QA}{QH}=\frac{TA}{HF}$ , $\frac{HF}{DH}=\frac{\cos A}{\cos C}$ Note that $\angle HTF=90-\angle C$ , by sines law in the triangle $TAH$ one can see that : $\frac{TA}{AH}=\frac{\sin (B-C)}{\cos C}$ , $P \iff \frac{TA}{DK}=\frac{TA}{DC-BD}=\frac{\cos A}{\cos C} \iff AH.\sin (B-C)=(DC-BD).\cos A$ $\cos A=\frac {AH}{2R_{ABC}} \implies P \iff \frac{DC-BD}{2R}=\cos C . \frac{AC}{2R}-\cos B . \frac{AB}{2R}=\cos C . \sin B-\cos B . \sin C=\sin(B-C)$ Which is true and we're done.

Cool problem! IMO 2/5 level i would say. Here is my solution:- Let $\omega_1$ be the circumcircle of $\Delta AFE$. Let the circle with diameter $AH$ intersect $\omega_1$ again at $I$. Claim: $I$ is the neccessary contact point. Proof: We first prove the circumcircle of $IPH$ or $\omega_2$ is tangent to $AH$. Just observe $\angle{AIH}=90$. Observe $\angle{ALP}=90-\angle{ABC}$ and $\angle{AIT}=90-\angle{ABC}$. at same time observe $\angle{FIH}=\angle{HAF}=90-\angle{ABC}$. This combines to give $\angle{TIF}=90$ so $ITPF$ , so $\angle{TIP}=\angle{ACB}$ further this shows so $\angle{PIH}=90-(90-\angle{ABC}+\angle{ACB})=\angle{ABC}-\angle{ACB}$ we can easily chase $\angle{AHP}$ which kills the proof. Seeing $\omega_2$ and $\omega_1$ are externally tangent is easy to see using $\angle{AIH}=90$ finishing the proof of the claim. To finally prove $HI$ passes through $K$ . Produce $AI$ to meet $\omega_2$ at $M$. its easy to see $M$ lies on $EF$ Claim: $MH$ is tangent to circumcircle of $AFE$. Proof: Basic angle chase Now this allows us to define $I$ completely in terms of $AFEH$. Now observe by more angle chase the proof we are looking for is same as $\Delta HCK \sim \Delta HIF$. as $\angle{HIF}=\angle{HCB}$. So all we wanna prove is $\frac{IF}{KC}=\frac{HI}{HC}$ As $I$ can be now defined in terms of $AFE$ basic sine rule chase computes $\frac{IF}{HI}$ finishes the problem,

wait what....... ofc inversion Do $-\sqrt{HA.HD}$ inversion. Let $X'$ be image of $X$ in inversion. we have $P'$ antipode of $H$ in $(BHC)$ and $L',T'$ are feet of perpendicular from $B,C$ to $HP'$. let $Y$ on $(BHC)$ such that $HY \parallel BC$. Observe that $\omega \leftrightarrow YP'$. Now as $H'$ is symmetric above perpendicular bisector of $BC$, we have $K \in YP'$. from $\angle BL'P' = \angle BKP' = 90$ we have $B,H,K,P'$ cyclic. hence $$\angle L'KB = \angle L'P'B = \angle HP'B = 90 - \angle B = \angle HLA = \angle L'T'D$$we get $L',T',D,K$ cyclic. Now as $\angle TAH = \angle HDL' = 90 - \angle B = 90 - \angle HDK$. Hence $H'K$ is tangent to $(HDK'T')$. Therefore we have $(ALT)$ is tangent to $\omega$ at $K'$. $K' \in KH$.

We $\sqrt{AH\cdot AD}$ invert such that $A,B,C\leftrightarrow D,E,F$, observe that $P$ goes to $P^*\equiv H-$antipode w.r.t $\odot(BHC)$ and $T,L$ go to $T^*,L^*\equiv$ the foot of perpendiculars from $B,C$ to $HP^*$. Since $\omega$ is tangent to $AD$ and $ABCP^*$ is a parallelogram we have $\omega \leftrightarrow P^*K$. So the problem is equivalent to proving Inverted problem wrote: Prove $\odot(DL^*T^*)$ is tangent to $P^*K. $ [asy][asy] import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -21.805291821757585, xmax = 18.29532968192365, ymin = -18.532610829960454, ymax = 14.046362467147715; /* image dimensions */ pen ccqqqq = rgb(0.8,0,0); draw((-1.4759731489746066,10.551691773887143)--(-8,-2)--(10,-2)--cycle, linewidth(0.7)); /* draw figures */ draw((-1.4759731489746066,10.551691773887143)--(-8,-2), linewidth(0.7)); draw((-8,-2)--(10,-2), linewidth(0.7)); draw((-8,-2)--(1.804248052564517,6.963993812466971), linewidth(0.7)); draw(circle((1,-5.293397053977596), 9.583656095413081), linewidth(0.7)); draw(circle((1,-2), 2.4759731489746075), linewidth(0.7) + red); draw(circle((-4.737986574487303,0.9824488329659772), 4.4199245049531335), linewidth(0.7)); draw((-4.1714020178639295,5.36590803429101)--(1,-2), linewidth(0.7)); draw((-1.475973148974607,3.964897665931955)--(3.475973148974605,-14.551691773887148), linewidth(0.7)); draw((3.4759731489746075,-2)--(3.475973148974605,-14.551691773887148), linewidth(0.7) + blue); draw((-1.4759731489746066,10.551691773887143)--(10,-2), linewidth(0.7)); draw((-1.475973148974607,3.964897665931955)--(3.4759731489746075,-2), linewidth(0.7)); draw((-8,-2)--(-0.4226947265315759,0.026421167857876002), linewidth(0.7)); draw((0.7787448240816803,-4.466067554138083)--(10,-2), linewidth(0.7)); /* dots and labels */ dot((-1.4759731489746066,10.551691773887143),linewidth(3pt) + dotstyle); label("$A$", (-1.2876597316387715,10.797366496261244), NE * labelscalefactor); dot((-8,-2),linewidth(3pt) + dotstyle); label("$B$", (-9.165362291185474,-2.9997396814758224), NE * labelscalefactor); dot((10,-2),linewidth(3pt) + dotstyle); label("$C$", (10.818188269472547,-2.287630975528103), NE * labelscalefactor); dot((-1.475973148974607,3.964897665931955),linewidth(3pt) + dotstyle); label("$H$", (-1.821741261099565,4.610922113340431), NE * labelscalefactor); dot((-1.4759731489746075,-2),linewidth(3pt) + dotstyle); label("$D$", (-2.088782025829962,-3.444807622693147), NE * labelscalefactor); dot((-4.1714020178639295,5.36590803429101),linewidth(3pt) + dotstyle); label("$F$", (-5.070737231986058,5.901619142870672), NE * labelscalefactor); dot((1.804248052564517,6.963993812466971),linewidth(3pt) + dotstyle); label("$E$", (2.0948566216129203,7.325836554766111), NE * labelscalefactor); dot((3.4759731489746075,-2),linewidth(3pt) + dotstyle); label("$K$", (3.6971012099953007,-1.6645358578238483), NE * labelscalefactor); dot((1,-2),linewidth(3pt) + dotstyle); label("$M$", (1.1602139450565316,-3.13326006384102), NE * labelscalefactor); dot((-0.4226947265315759,0.026421167857876002),linewidth(3pt) + dotstyle); label("$L'$", (0.3145848567436088,-0.596372798902269), NE * labelscalefactor); dot((0.7787448240816803,-4.466067554138083),linewidth(3pt) + dotstyle); label("$T'$", (-0.13048308447371906,-5.7146541229015035), NE * labelscalefactor); dot((3.475973148974605,-14.551691773887148),linewidth(3pt) + dotstyle); label("$P'$", (3.4745672393866367,-16.08473715326517), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] This is equivalent to proving $K\in \odot(M,D)$ as we already have $\angle P^*KM=90^{\circ}$. Now $DL^*\parallel BA$ because $\angle L^*DH=\angle HFL^*=\angle FEH=\angle DAB$. Since line through $M$ and midpoint of $BH$ is perpendicular bisector of $FB$ we also have $ML^*=MD=MK$ and similarly we also have $MT^*=MD=MK$ so $K\in \odot(DL^*KT^*)$.

We will prove that this point is $X$ which is $\overline{HK} \cap (HEF)$. You know the phrase which goes like ``it is so crazy that it actually works"? Well.... We are going to $\sqrt{bc}$ invert at $H$ in $\triangle HEF$! And so $\bullet$ $A$ is antipode of $H$ so $P$ swaps with $A$. $\bullet$ $\omega$ gets swapped with $\overline{A\infty_{HP}}$. $\bullet$ $L$ gets swapped with $\overline{HA} \cap (HPF)$ and $T$ with $\overline{HA} \cap (HPE)$. $\bullet$ See that $(BC;M\infty_{BC}) \overset{H}= (EF;Q_AH)=-1$ and so \[-1=(D,K;M,\infty_{BC}) \overset H= (A,X;Q_A,H) \overset{\sqrt{bc}}= (P,X^*;N,\infty_{EF})\]where $N$ is say midpoint of $\overline{EF}$. And now we will reform the labels because who the hell wants to work with these labels. Quote: Let $\triangle ABC$ be a triangle with $A'$ antipode; $D$ being foot of $A$-altitude; $K$ be point on $(ABC)$ such that $(AK;BC)=-1$; and finally let $L$ and $T$ be intersection points of $\overline{AD}$ with $(ABD)$ and $(ACD)$ respectively. Prove that line through $A'$ parallel to $\overline{AD}$ and $(LTD)$ are tangent at the reflection of $D$ across midpoint of $\overline{BC}$. The first point is obvious and we will prove that $(DTLX)$ is cyclic with diameter $\overline{DX}$ which makes the problem trivial. Here $X$ is reflection of $D$ across midpoint of $\overline{BC}$. So we need to prove $\measuredangle DLX=90^{\circ}$. $\bullet$ Let $A_1$ be point on $(ABC)$ such that $\overline{AA1} \parallel \overline{BC}$. $\bullet$ Let $Y=\overline{AA_1} \cap (ABD)$ and so $YADB$ is a rectangle. $\bullet$ Redefine $L$ as $\overline{AA'} \cap \overline{XY}$ and we need to prove that $L \in (ABD)$. And so see that \[\measuredangle LYB=\measuredangle XA_1B=\measuredangle CAD=\measuredangle A'AB=\measuredangle LAB\]And done. Remark: This problem is so good! There are a lot of solutions on AOPS that uses $-\sqrt{HA \cdot HD}$ inversion, but this inversion I will admit is so good. The reasons are that $\bullet$ When is the last time you really saw this type of an inversion. $\bullet$ And in the other solutions, after the inversion there is still some work to do; but I would say my solution is a bit more elegant as like the problem is almost immediately trivial once you do this inversion! Aight enough yap, bye!