Rough Sketch:(will add in full solution later) We claim that the circles are tangent at $A$ Claim 1: If $S$ is the Queue point then $ARST$ is cyclic Claim 2: $APEF$ is cyclic Then conclude.

Solved with help from ApraTrip, crazyeyemoody907, and trigadd123. [asy][asy] //igo prob //setup size(13cm); defaultpen(fontsize(10pt)); pen blu,grn,blu1,blu2,lightpurple; blu=RGB(102,153,255); grn=RGB(0,204,0); blu1=RGB(233,242,255); blu2=RGB(212,227,255); lightpurple=RGB(234,218,255);// blu1 lighter //defn int i=0; pair O,A,B,C,L,E1,F1,D,X1,Y1,X,K,H1,H,O1,P,M1,Q,N9,P1,R1,T1,U,T,R,E,F,Y,G,D1,S,J; O=(0,0); A=dir(123); B=dir(207); C=dir(333); L=(B+C)/2; E1=(A+B)/2; F1=(A+C)/2; D=intersectionpoint(B--C,A--(A+10(L-O))); X1=intersectionpoint(circumcircle(D,E1,F1),B--E1+0.1(B-E1)); Y1=intersectionpoint(circumcircle(D,E1,F1),A--F1+0.1(A-F1)); X=intersectionpoint(C--B+2(B-C),Y1--X1+100(X1-Y1)); H1=intersectionpoint(circumcircle(D,E1,F1),A--A+L-O); H=2H1-A; O1=circumcenter(B,O,C); P=foot(D,X,O1); M1=extension(P,D,E1,F1); Q=2M1-D; N9=(H1+L)/2; P1=intersectionpoint(circumcircle(D,E1,F1),M1--Q); R1=(P1+L)/2; T1=R1+(R1-N9)*(distance(P1,R1)/distance(N9,R1))^2; U=intersectionpoint(O1--N9,X--T1); T=intersectionpoint(X--U,circumcircle(D,E1,F1)); R=intersectionpoint(X--2U-X,circumcircle(D,E1,F1)); F=intersectionpoint(D--R+10(R-D),F1--F1+10(F1-E1)); E=intersectionpoint(D--T+10(T-D),E1--E1+10(E1-F1)); Y=intersectionpoint(P--D+5(D-P),L--(-10L)); K=intersectionpoint(C--T,B--R); G=intersectionpoint(A--X,L--3H-2L); D1=foot(D,A,X); S=intersectionpoint(O--H1,O1--2*N9-O1); J=intersectionpoint(O1--X,A--2D-A); //draw filldraw(A--B--C--cycle,blu1,blu); draw(circumcircle(D,E1,F1),blu+dotted); draw(circumcircle(B,O,C),blu+dotted); draw(circumcircle(B,O1,C),blu+dotted); draw(circumcircle(A,E,F),purple+dashed); draw(circumcircle(A,T,R),purple+dashed); draw(X--B^^L--G^^A--X,blu); draw(B--R^^C--T^^O--Q^^H1--J,blu+dotted); draw(E--D--F--cycle,magenta); draw(O1--X^^P--Q,purple); draw(A--D1--P--cycle,red); draw(O--O1--2*N9-O1--H1--cycle,blu); //label void pt(string s,pair P,pair v, pen a){filldraw(circle(P,.008),a,linewidth(.3)); label(s,P,v);} string labels[]={"$A$", "$B$", "$C$", "$O$", "$T$", "$R$", "$E$", "$F$", "$G$", "$X$", "$P$", "$O'$", "$K$", "$Q$", "$H$", "$H'$", "$S$", "$M'$", "$L$", "$J$", "$Y$", "$D'$", "$D$"}; pair points[]={A,B,C,O, T,R,E,F, G,X,P,O1, K,Q,H,H1, S,M1,L,J, Y,D1,D}; real dirs[]={135,240,0,0, 240,300,180,0, 150,180,270,270, 0,90,60,45, 45,45,300,270, 220,180,300}; pen colors[]={blu,blu,blu,blu, magenta,magenta,magenta,magenta, red,purple,purple,purple, purple,purple,blu,blu, blu,purple,blu,purple, purple, red,purple}; for (i=0; i<23; ++i) pt(labels[i],points[i],dir(dirs[i]),colors[i]); path clip[]; clip=circle((0,0),(distance(O,A))*1.60); clip(currentpicture,clip); [/asy][/asy] Let $X$ be defined on $BC$ such that $(X, D; B, C) = -1$; in other words, $X$ is collinear with the feet of the altitudes from $B$, $C$ in $\triangle{ABC}$ and also lies on line $AG$ if $G$ is the $A$-Orthocenter Miquel point in $\triangle{ABC}$. Let $O’$ be the center of $(BOC)$. Claim 1: $P$ is the projection of $D$ onto $O’X$. Proof: If $Y = DK \cap OO’$, then $P = YD \cap (YBC)$. Since $Y$ and the midpoint $L$ of $BC$ are inverses wrt $(BOC)$, $X$ lies on the polar of $Y$ so $Y$ lies on the polar of $X$, and so does $K$ by Brocard on $BTRC$, so $DK$ is the polar of $X$, meaning that $P$ is the inverse of $X$ wrt $(BOC)$, or the projection of $D$ onto $O’X$. Claim 2: $AEPF$ is cyclic. Proof: By this problem https://artofproblemsolving.com/community/c6h3196286_a_difficultly_problem, $PD$ bisects $EF$, and if this midpoint is $M’$ then $M’D \cdot M’P = M’E^2 = M’F^2$, so by Power of a Point at $M’$, the reflection $Q$ of $D$ over the midpoint of $EF$ lies on $(PEF)$. $AQFE$ is a cyclic isosceles trapezoid, so $AEPF$ is cyclic and in fact harmonic too. Claim 3: The projection $D’$ of $D$ onto $AX$ lies on $AEPF$. Proof: $DD’XP$ is cyclic, so $\angle{AD’P} = \angle{XDQ} = 180^{\circ} - \angle{AQP} \implies AD’PQ$ is cyclic. Now we just have to show that $(ART), (AD’P)$ are tangent at $A$. Claim 4: The $A$-Orthocenter Miquel Point, $G$, lies on $(ART)$. Proof: Note that $X \in RT$ since $Pow_{(BOC)}(X) = XB \cdot XC = XG \cdot XA = XD \cdot XL = Pow_{(DMN)}(X)$. Thus, $XR \cdot XT = XG \cdot XA \implies AGTR$ is cyclic. Let $N_9$ be the nine-point center of $\triangle{ABC}$, and the midpoint of $OH$, and let $H’$ be the midpoint of $AH$. Then $OH’$ and $N_9O’$ are the perpendicular bisectors of $AG, RT$ so the center of $(ART)$ is $S = OH’ \cap N_9O’$. Claim 5: $(ART), (AD’P)$ are tangent if and only if $\angle{APD} = \angle{DAS}$. Proof: The circles are tangent if and only if the center of $(AD’P)$ lies on $AS$, which happens if and only if $\angle{D’AS} = 90^{\circ} - \angle{D’PA}$. By cyclic quadrilateral $DD’XP$, $\angle{D’PX} = \angle{D’DX} = \angle{D’AP}$, so they are tangent if and only if $\angle{APD} = \angle{DAS}$. *cue start of length bash Let $J = O'X \cap AD$. $\angle{APD} = \angle{DAS} \iff \tan \angle{APD} = \tan \angle{DAS}$. If $Z \in XP$ such that $\angle{ZAD} = 90^{\circ}$, then $ZADP$ is cyclic $$\implies \tan \angle{APD} = \tan \angle{AZD} = \frac{AD}{AZ} = \frac{AD}{XD \cdot \frac{JA}{JD}}.$$Since $N_9$ is the midpoint of $OH$, $$H’S : SO = O’L : O’O \implies \tan \angle{DAS} = \frac{DL}{AD + OL\left(\frac{O’O}{O’L}-1\right)} = \frac{DL}{AD + \frac{OL^2}{O’L}}$$so we want to show that $$\frac{AD}{XD \cdot \frac{JA}{JD}} = \frac{DL}{AD + \frac{OL^2}{O’L}}.$$Since $H$ is the orthocenter of $\triangle{AXL}$, $\frac{HD}{XD} = \frac{DL}{AD}$, so this simplifies to $$\frac{HD}{AD + \frac{OL^2}{O’L}} = \frac{JD}{JA}.$$So, it suffices to show that $$JD = \frac{AD \cdot HD}{\frac{OL^2}{O’L} + 2 \cdot OL} = \frac{AD \cdot HD}{\frac{OL}{O’L}(OL + 2 \cdot O’L)} = \frac{AD \cdot HD}{\frac{OL}{O’L} \cdot WL}$$if $W$ is the $O$-antipode in $(BOC)$. $(X, D; B, C) = -1 \implies LX \cdot LD = LB^2$, so $$\frac{AD \cdot HD \cdot O’L}{OL \cdot WL} = O’L \cdot \frac{Pow_{(ABC)}(D)}{LB^2} = O’L \cdot \frac{Pow_{(ABC)}}{XL \cdot DL} = O’L \cdot \frac{DL \cdot DX}{XL \cdot DL} = O’L \cdot \frac{DX}{XL} = JD,$$so $\tan \angle{APD} = \tan \angle{DAS} \implies \angle{APD} = \angle{DAS}$ and we are done. $\square$

Solved with help from Infinityfun and erkosfobiladol. Let $X=RT\cap BC$. Note that $(X,D;B,C)=-1$ and $PD$ is the angle bisector of $\measuredangle BPC$ thus, $\measuredangle APD=90$. Take $\sqrt{\frac{bc}{2}}$ inversion and reflect over the angle bisector of $\measuredangle A$. We will prove that $P\in (AEF)$ and $(AEF)$ is tangent to $(ART)$. New Problem Statement: Let $ABC$ be a triangle circumcenter $O$ and midpoints of $AC,AB$ are $M,N$ respectively. $(BOC)$ meets nine point circle of $ABC$ at $R,T$. $(AMT)$ and $(ANR)$ intersect at $K$. $X$ is the intersection of $(AMNO)$ with $(ART)$. $P$ is taken on $(AKO)$ such that $AO$ and $(XPO)$ are tangent to each other. $(AOR)$ and $(AOT)$ meet $(ABC)$ at $E,F$ respectively. Prove that $P,E,F$ are collinear and $\overline{PEF}\parallel RT$. Invert around $(ABC)$. New Problem Statement: $ABC$ is a triangle where its incircle is tangent to $BC,CA,AB$ at $D,E,F$ respectively. $EF$ meets $(ABC)$ at $R,T$. $DR,DT$ intersects the incricle at $U,V$. $K$ is the intersection of $(BDT)$ with $(CDR)$. $(DRT)$ meets $BC$ at $X$. $P$ is taken on $KD$ such that $PX\perp BC$. Prove that $P,U,V,I$ are concyclic and this circle is tangent to $(IRT)$. Claim: $X$ is the midpoint of $BC$. Proof: Let $EF\cap BC=W$. Since $(W,D;B,C)=-1,$ we have \[WB(WB+BC)\overset{?}{=}(WB+BD)(WB+\frac{BC}{2})\iff \frac{1}{2}.WB.BC\overset{?}{=}BD.WB+\frac{BC.BD}{2}\]\[WB(DC-DB)=WB(BC-2DB)\overset{?}{=}BD.BC\iff WB.DC\overset{?}{=}WC.BD\]which is true.$\square$ We want to prove that there exists a point $P$ which hold $PB=PC,P\in KD,P\in (IUV)$ and $(IUV)$ is tangent to $(IRT)$. Claim: $ABC$ is a triangle and its incircle is tangent to $BC,CA,AB$ at $D,E,F$ respectively. $EF$ intersects $(ABC)$ at $R,T$. $DR,DT$ intersects the incircle at $U,V$. If $J$ is the intersection of tangents to $(ABC)$ at $B,C,$ then show that $JU,JV$ are tangent to the incircle. Proof: Let the tangents to the incircle from $R,T$ intersect $(ABC)$ at $R_1,R_2,T_1,T_2$. By Poncelet Porism, $R_1R_2$ is tangent to incircle. By DDIT, $(\overline{RR_1},\overline{RR_2}),(\overline{RE},\overline{RB}),(\overline{RA},\overline{RC})$ are involution pairs so $R_1R_2,AC,TB$ are concurrent. Let $R_3$ be the concurrency point. By DDIT, $(\overline{RR_1},\overline{RR_2}),(\overline{RF},\overline{RC}),(\overline{RA},\overline{RB})$ are involution pairs thus, similarily $AB,TC,R_1R_2$ are concurrent. Let this point be $R_4$. We apply Pascal on $CCABBT$ to get that $J,R_3,R_4$ are collinear. Hence $J\in R_1R_2$. We can conclude same thing for $T$ so $J\in T_1T_2$. Let $R_1R_2$ be tangent to incircle at $V'$. By DDIT, $(\overline{TV'},\overline{TR}),(\overline{TR_1},\overline{TR_2}),(\overline{TT_1},\overline{TT_2})$ are involution pairs thus, if we let $TV'\cap (ABC)=S,$ then $SR,R_1R_2,T_1T_2$ are involution pairs. Since $J=R_1R_2\cap T_1T_2$ and $R_1,R_2,T_1,T_2,S,R$ are concyclic, we verify that $R,S,J$ are collinear. \[(TR,TD;TB,TC)=-1=(TR,TS;TB,TC)\]Hence $T,D,S$ are collinear which yields $V=V'$. So $R_1R_2$ is tangent to incircle at $V$ and $T_1T_2$ is tangent to incircle at $U$.$\square$ Claim: $(IRT)$ is tangent to $(IUV)$. Proof: Let $IE,IF$ intersect $(IAC),(IAB)$ at $Y,Z$ respectively. $FI.FY=FA.FB=FR.FT$ so $Y\in (IRT)$. Similarily $Z\in (IRT)$. Let $O$ be the circumcenter of $(IRTYZ)$. By previous claim, we get that $JI$ is the diameter of $(IUV)$ so proving the collinearity of $J,I,O$ would be enough. Let $M_B,M_C$ be the intersection of $BI,CI$ with $(ABC)$. $OM_C\perp IF\perp AB$ and $OM_B\perp IE\perp AC$ hence $O$ is the intersection of the tangents to $(ABC)$ at $M_B,M_C$. Pascal at $BBM_BCCM_C$ and $M_BM_BBM_CM_CC$ imply $J,I,M_BC\cap BM_C$ and $O,I,BM_B\cap CM_C$ are collinear. Thus, $O,I,J$ are collinear.$\square$ Claim: $J\in KD$. Proof: Let $TD,RD$ intersect $(ABC)$ at $G,H$. Since $(R,G;C,B)=(TR,TG;TC,TB)=-1$ we get that $R,G,J$ are collinear. Similarily $T,G,J$ are collinear. Since $RT\cap GH$ lies on the polar line of $J$ with respect to $(RTGH),$ we observe that $RT,GH,BC$ are concurrent. Let $W$ be the concurrency point. Let $BH\cap CG=L$. $(W,D;B,C)=-1=(JW,JL;JB,JC)$ hence $J,L,D$ are collinear. Inverting from $D$ with radius $\sqrt{-DB.DC}$ yields $DJ$ passes through the second intersection of $(BDT)$ and $(CDR)$ which is $K$ as desired.$\blacksquare$

Attachments: