Invert from $H$ with radius $-HA.HD$. Then radical axis on $(ARS),(ABC),(BHC)$. Finish by PoP.

Let the tangent from $A$ to $(ABC)$ intersect $BC$ at $K$. Since $EF$ is an anti parallel, we get $AK//EF$ so $AK\perp AT$. Then, $|KD|\perp |KT|=|KA|^2=|KB|\cdot |KC|=|KS|\cdot |KX|$ where $X=KS\cap (BHC)$. Hence, the question is equivalent to proving the collinearity of the points $K, S, R$. Let $EF\cap BC=L, LH\cap (BHC)=M$ and $DH\cap (BHC)=N, DH\cap EF= W$. We know the point $N$ is the reflection of $A$ over $BC$. Yields, $|KN|^2=|KA|^2=|KB|\cdot |KC|$, so $KN$ is tangent to $(BHC)$. As $(M, N, B, C)=(HM, HN, HB, HC)=(L, D, B, C)=-1$, we find that $KM$ is also tangent to $(HBC)$. Then, in order to prove that the points $K, S, R$ are collinear, we must prove that $-1 \overset{?}{=}(S, R, M, N)= (HS, HR, HM, HN)=(P, Q, L, W)$. Since the circles $(DEF)$ and $(DPQ)$ are tangent, we find that $\angle PDF=\angle EDQ$. Also, $DW$ is the angle bisector of $\angle EDF$ so we get $\angle PDW=\angle WDQ$. Also, $\angle WDL=90^{\circ}$ so the points $L, P, W, Q$ indeed form a harmonic bundle.

Let $AD\cap EF=X$ and $EF\cap BC=K$. Since $\angle KDX=90^{\circ}$ and $\angle EDX=\angle FDX$, we have $(K,X;F,E)=-1$. Since circles $(DEF)$ and $(DPQ)$ are tangent, we have $\angle PDX=\angle FDX-\angle FDP=\angle EDX-(\angle PQD-\angle FED)=\angle QDX$, so $(K,X;P,Q)=-1$. Let $(ABC)=\omega_1$, $(BHC)=\omega_2$, $XH\cap \omega_2=Z$, $KH\cap \omega_2=N$ and $RS\cap BC=L$. We have $(K,X;F,E)\overset{H,\omega_2}{=} (N,Z;C,B)=-1$ and $(K,X;P,Q)\overset{H,\omega_2}{=} (N,Z;R,S)=-1$, hence tangent from $N$ and $Z$ to $\omega_2$ intersect on $BC$ and $RS$. That means $LZ$ is tangent to $\omega_2$ and $LZ^2=LA^2=LR\cdot LS$. Since $\angle LAT=90^{\circ}$ and $LA^2=LD\cdot LT=LR\cdot LS$, we have $R,S,D,T$ cyclic. [asy][asy] size(20cm); pair A = dir(120), B = dir(215), C = dir(325); pair D = foot(A, B, C), E = foot(B, A, C), F = foot(C, A, B); pair O = circumcenter(A, B, C); pair H = orthocenter(A, B, C); pair P = 0.85 * F + 0.15 * E; pair Q = extension(extension(A, P, B, C), H, E, F); pair R = intersectionpoints(P -- (21 * H - 20 * P), circumcircle(B, H, C))[1]; pair S = intersectionpoints(Q -- (21 * H - 20 * Q), circumcircle(B, H, C))[1]; pair T = extension(A, O, B, C); pair Z = 2 * D - A; pair K = extension(B, C, E, F); pair L = extension(B, C, R, S); pair X = extension(E, F, A, D); pair N = intersectionpoints(H -- (2 * H - K), circumcircle(B, H, C))[1]; draw(A -- B -- C -- cycle); draw(A -- D ^^ B -- E ^^ C -- F); draw(E -- F); draw(R -- P ^^ S -- Q); draw(A -- Z); draw(K -- E ^^ A -- L -- R ^^ B -- L); draw(K -- N); draw(A -- T); draw(unitcircle); draw(circumcircle(B, H, C)); draw(circumcircle(D, E, F)); draw(circumcircle(D, P, Q)); draw(circumcircle(R, T, S), dashed); dot("$A$", A, dir(120)); dot("$B$", B, dir(270)); dot("$C$", C, dir(0)); dot("$D$", D, dir(300)); dot("$E$", E, dir(70)); dot("$F$", F, dir(140)); dot("$H$", H, dir(0)); dot("$O$", O, dir(270)); dot("$P$", P, dir(90)); dot("$Q$", Q, dir(90)); dot("$R$", R, dir(315)); dot("$S$", S, dir(225)); dot("$T$", T, dir(270)); dot("$Z$", Z, dir(270)); dot("$K$", K, dir(90)); dot("$L$", L, dir(150)); dot("$X$", X, dir(340)); dot("$N$", N, dir(90)); [/asy][/asy]

Lot of inversion approaches on this problem. I enjoyed this problem a lot! We perform a $\sqrt{AD \cdot DH}$ inversion centered at $D$ followed by a reflection across the $A-$altitude. Then the problem rewrites to the following. Inverted Problem wrote: Let $\triangle ABC$ be an acute triangle, where $H$ is the orthocenter and $D,E,F$ are the feet of the altitudes from $A,B,C$ respectively. A line parallel to $\overline{EF}$ intersects $(DEF)$ at points $P$ and $Q$. Let $R$ and $S$ be the second intersections of circles $(APD)$ and $(AQD)$ with $(ABC)$. Let $T$ be the intersection of $(XHD)$ and $\overline{BC}$ where $X$ is the second intersection of line $\overline{HM}$ with $(DEF)$. Show that points $R$ , $S$ and $T$ are collinear. The following observation is our key claim. Claim : Points $P$ , $R$ , $H$ and $Q$ , $S$ , $H$ are collinear. Proof : Let $R' = \overline{PH} \cap (ABC)$ , $P' = \overline{PH} \cap (DEF)$ and $N = \overline{AH} \cap (DEF)$. Since it is well known that the nine-point circle is the image of the circumcircle under homothety at $H$ with scale factor $\frac{1}{2}$, it follows that $P'$ is the midpoint of segment $HR'$ and $N$ is the midpoint of segment $AN$. Then, \[HR'\cdot HP = 2HP' \cdot HP = 2HN \cdot HD = HA \cdot HA\]which implies that $APDR'$ is cyclic, and thus $R'\equiv R$ proving the claim. Claim : Line $\overline{TH}$ is tangent to $(BHC)$ at $H$. Proof : First note that since $\measuredangle HXT = \measuredangle HDT = 90^\circ$, points $T$ , $X$ and $N$ are collinear. So, \begin{align*} \measuredangle DHT &= \measuredangle DXT \\ & = \measuredangle DXN\\ &= \measuredangle DFN \\ &= \measuredangle FDN + \measuredangle DNF \\ &= 90 + \measuredangle FDB + 2 \measuredangle AEF\\ &= 90 + \measuredangle BAC + 2\measuredangle CBA\\ &= 90 + \measuredangle CBA + \measuredangle BCA\\ &= \measuredangle DAF + \measuredangle AFE\\ &= \measuredangle (EF;AD) \end{align*}which implies that $TH \parallel EF$. But then, \[\measuredangle BHT = \measuredangle BEF = \measuredangle BCH\]which proves the claim. Further note that, \[HP \cdot HR = HA \cdot HD = HQ \cdot HS\]so quadrilateral $PQSR$ must be cyclic. Now we can finish off. Claim : Circles $(BHC)$ and $(RHS)$ are tangent to each other at $H$. Proof : This is simply an angle chase. Note that letting $Y=EF \cap QH$ we have, \begin{align*} \measuredangle SHB & = \measuredangle QHE \\ &= \measuredangle QYE + \measuredangle YEH \\ &= \measuredangle HQP + \measuredangle FEH \\ &= \measuredangle SQP + \measuredangle FEB \\ &= \measuredangle SRP + \measuredangle FCB \\ &= \measuredangle SRH + \measuredangle HCB \end{align*}which proves the claim. But the tangent to $(BHC)$ at $H$ is simply $\overline{HT}$. By Radical Center Theorem on circles $(ABC)$ , $(BHC)$ and $(RHS)$ we have that lines $\overline{BC}$ , $\overline{RS}$ and $\overline{TH}$ must concur which implies that point $R$ , $S$ and $T$ are collinear, as desired.

In the orthic triangle $DEF$ of $ABC$, $H$ is the incenter. (Proof: $\angle EFH = \angle EAH = \angle DBH = \angle DFH$.) Let the intersection of $EF$ and $AD$ be $A'$. In the triangle $DFA'$, $FH$ is the internal angle bisector, and $FA$ is the external angle bisector: $$\frac{AA'}{AD} = \frac{HA'}{HD}.$$ Rewriting the same ratio: $$\frac{HA - HA'}{AD} = \frac{HA'}{AD - HA}.$$ Expanding: $$HA \cdot AD - HA^2 - HA' \cdot AD + HA \cdot HA' = AD \cdot HA'.$$ From this: $$2 \cdot AD \cdot HA' = HA(AD - HA + HA') = HA(DH + HA') = HA \cdot DA'.$$ We deduce: $$\frac{HA'}{HA} = \frac{1}{2} \cdot \frac{DA'}{DA} \qquad (1)$$ --- Let $EF$ intersect $BC$ at $L$. The tangent at $D$ to the circle $(DEF)$ intersects $EF$ at $M$. We have: $$\angle A'DE = 90^\circ - \angle BAC.$$$$\angle EDM = \angle DFE = 180^\circ - 2\angle BCA.$$\[ \begin{array}{lcl} \angle MDL &=& 90^\circ - (\angle A'DE + \angle EDM) \\ &=& \angle BAC + 2\angle BCA - 180^\circ \\ &=& \angle BCA - \angle ABC \\ &=& \angle BCA - \angle AEF \\ &=& \angle BCA - \angle LEC \\ &=& \angle DLM. \end{array} \]Therefore, $$ML = MD = A'M \qquad (2)$$ --- Let the parallel to $EF$ passing through $A$ intersect $CF$, $BE$, and $BC$ at points $F'$, $E'$, and $K$, respectively. From $(1)$, $$\frac{HA'}{HA} = \frac{1}{2} \cdot \frac{DA'}{DA} = \frac{1}{2} \cdot \frac{A'L}{AK} = \frac{A'M}{AK},$$so $H$, $M$, and $K$ are collinear. Since $AK \parallel EF$ and $AT \perp EF$, we have $AT \perp AK$. By Euclid, $$KA^2 = KD \cdot KT \qquad (3)$$ As $\angle ABC = \angle AEF = \angle EAK$, the line $KA$ is tangent to the circumcircle of $(ABC)$. $$KA^2 = KC \cdot KB \qquad (4)$$ Because $\angle HBC = \angle EFH = \angle E'F'H$, $$KE' \cdot KF' = KC \cdot KB = KA^2 \qquad (5).$$ --- Since the radical axis of circles $(DEF)$ and $(DPQ)$ is $DM$, $$ME \cdot MF = MP \cdot MQ = MD^2 = MA'^2 \qquad (6)$$ Let $HP$ and $HQ$ intersect line $AK$ at $P'$ and $Q'$, respectively. From the similarity $\triangle HMF \sim \triangle HKF'$, the equality in $(6)$ implies $$KP' \cdot KQ' = KE' \cdot KF' = KA^2 \qquad (7)$$ Combining $(7)$ and $(4)$, the points $P'$, $Q'$, $B$, and $C$ are concyclic. --- We will now show that the points $P$, $Q$, $R$, and $S$ are concyclic. Depending on whether $P$ and $Q$ lie on or outside segment $EF$, the proof slightly differs. Let us proceed based on the configuration in the diagram. \[ \angle HRS = \angle HBS = \angle HBC - \angle SBC = \angle EFH - \angle SHC = \angle EFH - \angle FHQ = \angle PQH. \]This implies that $P$, $Q$, $R$, and $S$ are concyclic. Since $PQ \parallel P'Q'$, the points $P'$, $Q'$, $R$, and $S$ are also concyclic. --- The radical axes of the circles $(P'Q'RS)$, $(P'Q'BC)$, and $(BRSC)$ intersect at a single point (the radical center of the three circles). This point is the intersection of $P'Q'$ and $BC$, which is $K$. Therefore, $R$, $S$, and $K$ are collinear. Finally, since $$KS \cdot KR = KC \cdot KB = KA^2 = KD \cdot KT,$$the points $R$, $S$, $D$, and $T$ are concyclic.

Attachments:

Let $\omega_1 = (DEF)$, $\omega_2 = (DPQ)$, and $\omega_3 = (BHC)$. Since $\angle BFC = \angle BEC$, we have $(BCEF) = \omega_4$. $\angle EFH = \angle EBC = \angle EBS + \angle SBC = \angle HRS + \angle SHC$ $\angle EFH = \angle FQH + \angle QHF = \angle FQH + \angle SHC = \angle HRS + \angle SHC$, so $\angle FQH = \angle HRS$. Therefore, $\omega_5 = (PQRS)$. Since $\angle FQH = \angle HRS = \angle SCH$, the points $S, C, Q, F$ are concyclic. In this case, $QH \cdot HS = FH \cdot HC$. Since $QH \cdot HS = PH \cdot HR$ and $EH \cdot HB = FH \cdot HC$, the power of point $H$ with respect to the circles $\omega_4$ and $\omega_5$ is equal. Hence, $H$ lies on the radical axis of these two circles. The radical axis of $\omega_3$ and $\omega_4$ is $BC$. The radical axis of $\omega_3$ and $\omega_5$ is $RS$. Let the intersection of $BC$ and $RS$ be $K$. The radical axis of $\omega_4$ and $\omega_5$ will be $HK$. Let $EF$ and $HK$ intersect at $M$. Since $M$ lies on the radical axis of $\omega_4$ and $\omega_5$, its power with respect to these two circles is equal, meaning $ME \cdot MF = MP \cdot MQ$. Since the power of point $M$ with respect to $\omega_1$ is $ME \cdot MF$, and its power with respect to $\omega_2$ is $MP \cdot MQ$, point $M$ lies on the radical axis of $\omega_1$ and $\omega_2$. This is the line tangent to these two circles at point $D$. Therefore, $DM$ is tangent to $\omega_1$. Let $FM$ intersect $BC$ and $AD$ at points $L$ and $A'$, respectively. $\angle ADE = \angle FCA = 90^\circ - \angle A$. $\angle EFD = 180^\circ - \angle AFE - \angle DFB = 180^\circ - 2\angle C$. By the Tangent-Secant angle theorem, $\angle EDM = \angle EFD = 180^\circ - 2\angle C$. $\angle MDL = 90^\circ - \angle ADE - \angle EDM = 90^\circ - (90^\circ - \angle A) - (180^\circ - 2 \angle C) = \angle C - \angle B$. $\angle MLD = \angle ACB - \angle LEC = \angle C - \angle AEF = \angle C - \angle B$. Thus, $MD = ML = MA'$. Let $AA' = x$, $A'H = y$, $HD = z$, $DL = p$, and $LK = q$. Let a line parallel to $AD$ through $L$ intersect $HK$ at $L'$. Then $LL' = A'H = y$. $\frac{DL}{DK} = \frac{p}{p+q}$. $\frac{DL}{DK} = \frac{HD - LL'}{HD} = \frac{z-y}{z}$. In $\triangle A'FD$, since $\angle A'FH = \angle EAH = \angle DBH = \angle DFH$, $FH$ is the internal angle bisector. Since $\angle AFH = 90^\circ$, $FA$ is the external angle bisector. Thus, $\frac{AA'}{AD} = \frac{HA'}{HD} = \frac{x}{x+y+z} = \frac{y}{z}$. Rewriting, $\frac{z-y}{z} = \frac{x+y+z - x}{x+y+z} = \frac{y+z}{x+y+z} = \frac{DA'}{DA}$. Therefore, $\frac{DL}{DK} = \frac{DA'}{DA}$, which means $A'L \parallel AK$. This implies that since $TA \perp EF$, $TA \perp KA$. By Euclid’s theorem, $KA^2 = KD \cdot KT$. Since $\angle KAE = \angle AEF = \angle ABC$, the line $KA$ is tangent to the circumcircle $(ABC)$ at $K$. Thus, $KA^2 = KC \cdot KB$. Since the power of point $K$ with respect to $\omega_4$ is $KC \cdot KB$ and with respect to $\omega_5$ is $KS \cdot KR$, we have $KS \cdot KR = KC \cdot KB = KA^2 = KD \cdot KT$. Thus, the points $S, R, D, T$ are concyclic.

Attachments:

I think this is easier than p1 because inverting through $H$ is very intuitive. Consider the inversion with center $H$ and radius $-\sqrt{HA.HD}$. Now notice that $EF$ is sent to $BHC$ so $P$ is sent to $R$ and $Q$ is sent to $S$ and obviosly $D$ is sent to $A$. So it's enough to prove $AT'PQ$ is cyclic. Now by easy angle chasing we get that $AT'FE$ cyclic. So it suffices to prove that $T'A$, $FE$ and the tangent of $(DEF)$ at $D$ are concurrent. Which implies proving that $(T'AD)$ and $(DEF)$ are tangent. When we invert this is equivalent to proving $(ADT)$ and $(ABC)$ are tangent. Now by easy angle chasing we get that $AH$ and $AT$ are isogonal conjugates. Which means $A$,$O$,$T$ collinear where $O$ is the circumcenter of $ABC$. Now since the circumcenter of $ADT$ lies on $AT$ because it's a right triangle we get that $(ADT)$ and $(ABC)$ are tangent. So let's call the intersection of $T'A$, $EF$, and the tangent $X$. We get that $XT'.XA = XF.XE = XD^2 = XP.XQ$ so done.

First I'll invert around $D$ with arbitrary radius. We get that this problem is equivalent to the following problem: Triangle $\triangle ABC$ with orthocentre $H$ and altitudes $AD, BE, CF$. $P, Q$ are on $(DEF)$ with $PQ\parallel EF$. $R=(ADP)\cap (ABC)$ and $S=(ADQ)\cap (ABC)$. $T$ is on $BC$ with $\angle DTH=|\angle DCH-\angle DBH|$. Prove that $R, S, T$ are collinear. Claim: $S, H, Q$ are collinear. Proof: I'll use the phantom point $S'$ defined as $S'=QH\cap (ADQ)$. Then by PoP from $H$ we have $HA\cdot HD=HB\cdot HE=HC\cdot HF=HQ\cdot HS'$. So by inversion around $H$ with negative radius and using that $Q$ is on $(DEF)$ we get that $S$ is on $(ABC)$, which is enough to show $S'=S$. This finishes the claim. Analogously $R, H, P$ are collinear. Now consider the inversion around $H$ with radius $-\sqrt{HA\cdot HD}$. It sends $B\leftrightarrow E$, $C\leftrightarrow F$, $P\leftrightarrow R$, $Q\leftrightarrow S$. $T$ gets sent to the point on $(EFH)$ such that $AT, AH$ are isogonal in $\angle BAC$. We wish to show that $T$ is on $(PQH)$. By $AT, AH$ isogonal we get $AT\perp EF$. By Thales we have $AT\perp TH$ and combining gives $TH\parallel EF\parallel PQ$. Now radax theorem on $(AEFH), (DPFEQ), (DPHQ)$ implies $T$ is on $(DPHQ)$ which finishes the proof.

Let $\omega_1 = (DEF)$, $\omega_2 = (DPQ)$, and $\omega_3 = (BHC)$. Since $\angle BFC = \angle BEC$, we have $(BCEF) = \omega_4$. As $\angle EFH = \angle EBC = \angle EBS + \angle SBC = \angle HRS + \angle SHC$, and $\angle EFH = \angle FQH + \angle QHF = \angle FQH + \angle SHC = \angle HRS + \angle SHC$, it follows that $\angle FQH = \angle HRS$. Hence, $(PQRS) = \omega_5$. Let $(ADT) = \omega_6$ and $(ABC) = \omega_7$. Let $N$ be the midpoint of $AT$, and let $O$ be the center of $\omega_7$. Since $AT$ and $AD$ are isogonal conjugates, $T, O, N, A$ are collinear. The circles $\omega_6$ and $\omega_7$ are tangent at $A$. Let $EF$ and $BC$ intersect at $L$, and let $EF$ and $AD$ intersect at $A'$. Let $M$ be the midpoint of $A'L$. Since $\angle NDA = \angle NAD = 90^\circ - \angle DA'M = 90^\circ - \angle A'DM$, we have $ND \perp DM$. Thus, $DM$ is tangent to $\omega_6$. As $\angle MDE = 90^\circ - \angle ADE - \angle MDL = 90^\circ - (90^\circ - \angle A) - (\angle C - \angle B)$, we find $\angle MDE = \angle A + \angle B - \angle C = 180^\circ - 2\angle C = \angle EFD$. Therefore, $DM$ is also tangent to $\omega_1$. Since $\angle FQH = \angle HRS = \angle SCH$, the points $S, C, Q, F$ are concyclic. Thus, $QH \cdot HS = FH \cdot HC$. As $QH \cdot HS = PH \cdot HR$ and $EH \cdot HB = FH \cdot HC$, the power of $H$ with respect to $\omega_4$ and $\omega_5$ is equal. Hence, $H$ lies on the radical axis of these two circles. The power of $M$ with respect to $\omega_1$, $\omega_2$, and $\omega_6$ is equal, as $MD^2 = ME \cdot MF = MP \cdot MQ$. Therefore, the power of $M$ with respect to $\omega_4$ and $\omega_5$ is also equal, placing $M$ on their radical axis. Thus, the radical axis of $\omega_4$ and $\omega_5$ is $HM$. The line $HM$ is independent of the choice of $\omega_2$. Note that $\omega_6$ also belongs to the family of circles $\omega_2$. Hence, the power of $M$ with respect to $\omega_6$ and $\omega_4$ is equal. As $EH \cdot HB = FH \cdot HC = DH \cdot HA$, the power of $H$ with respect to $\omega_4$ and $\omega_6$ is also equal. Thus, the radical axis of $\omega_4$, $\omega_5$, and $\omega_6$ is $HM$. Let $HM$ and $BC$ intersect at $K$. The radical center of $\omega_4$, $\omega_6$, and $\omega_7$ is the point $K$. Hence, $KA \perp NT$. By Euclid's theorem, $KD \cdot KT = KA^2 = KS \cdot KR$, which implies that the points $R, S, D, T$ are concyclic.