Let $ H $ be the orthocenter of $ \triangle ABC $ and let $ T $ be the anti-steiner point of $ HP $ WRT $ \triangle ABC $. Since the second intersection $ X $ of $ AH $ and $ \odot (ABC) $ is the reflection of $ H $ in $ BC $, so $ D, $ $ T, $ $ X $ are collinear, hence from Reim's theorem $ \Longrightarrow $ $ T $ $ \in $ $ \odot (PDL) $. Similarly, we can prove $ T $ lies on $ \odot (PEM) $ and $ \odot (PFN) $, so these three circles have a common point $ T $ ($ \neq $ $ P $).

Consider inversion, it's easy to get the image of $D,E,F,$ but I don't know how to do next? My friend told me that the problem after inverting the diagram is equivalent to Sondat's Theorem! Can someone give me a source about that?

hmm jayme proposed this question on this site about a year ago! See Coaxal circles.

Let $H$ be orthocenter of $\bigtriangleup ABC$.Let $H_A,H_B,H_C$ reflections of $H$ over $BC,CA,AB$, it's well-known that they lie on $\odot ABC$. Then $H_AD, H_BE, H_CF$ are concurrent at Anti-Steiner point $S$ of $PH$ wrt. $\bigtriangleup ABC$, which lies on $\odot ABC$. $\angle PDS= 360^{\circ}-90^{\circ}-\angle BCS-\angle H_ASC=270^{\circ}-\angle BAS-90^{\circ}-\angle C=180^{\circ}-\angle BAS-\angle BCA=\angle ALS=\angle PLS$ $\rightarrow S\in \odot PDL$, similarily $S\in \odot PEM$ and $S\in \odot PFN$, hence $S$ is the desired point.

Hello. My solution. Let $\omega$ be the circumcircle of $\triangle{ABC}$ and $T\neq L$ be its intersection with the circumcircle of $\triangle{PDL}$. We will show that $PETM$ is cyclic.Let $Q,R$ be the second intersections of $AE,BD$ with $\omega$. [asy][asy]import graph; size(12cm); 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 = -4.640552165934572, xmax = 16.800836449463805, ymin = -5.262246015207001, ymax = 6.869065964557864; /* image dimensions */ pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen sqsqsq = rgb(0.12549019607843137,0.12549019607843137,0.12549019607843137); draw((1.5887633870344084,4.122397208144715)--(0.,0.)--(7.,0.)--cycle, aqaqaq); /* draw figures */ draw((1.5887633870344084,4.122397208144715)--(0.,0.), sqsqsq); draw((0.,0.)--(7.,0.), sqsqsq); draw((7.,0.)--(1.5887633870344084,4.122397208144715), sqsqsq); draw(circle((3.5,1.0184588855083412), 3.645169200664201)); draw(circle((5.754396295530562,0.), 2.944702954172458), linetype("2 2")); draw((2.9226598752830073,0.8078021630050068)--(2.9226598752830073,-0.8078021630050068)); draw((2.9226598752830073,0.8078021630050068)--(5.138560360236284,3.7164888395768223)); draw((-1.624597948619468,2.56030595564012)--(2.9226598752830073,0.8078021630050068)); draw((-0.12392909019013483,1.4113920232615658)--(7.,0.)); draw((1.5887633870344084,4.122397208144715)--(4.271575759791679,-2.5441145310506474)); draw((0.,0.)--(7.026234411480838,1.942000642419889)); draw((2.9226598752830073,0.8078021630050068)--(6.683310936129654,2.794350206898316)); draw((6.683310936129654,2.794350206898316)--(4.271575759791679,-2.5441145310506474)); draw((6.683310936129654,2.794350206898316)--(7.026234411480838,1.942000642419889)); draw((5.138560360236284,3.7164888395768223)--(7.026234411480838,1.942000642419889)); draw((2.9226598752830073,-0.8078021630050068)--(4.271575759791679,-2.5441145310506474)); draw((1.5887633870344084,4.122397208144715)--(7.026234411480838,1.942000642419889)); draw((1.5887633870344084,4.122397208144715)--(6.062591929471243,3.610828670951718)); draw((0.,0.)--(5.980167730329973,-1.6528753374785081)); draw((5.138560360236284,3.7164888395768223)--(7.,0.)); draw((7.,0.)--(2.9226598752830073,-0.8078021630050068)); /* dots and labels */ dot((1.5887633870344084,4.122397208144715),linewidth(2.pt) + dotstyle); label("$A$", (1.4654071860068831,4.289650792780616), NE * labelscalefactor); dot((0.,0.),linewidth(2.pt) + dotstyle); label("$B$", (-0.42885083076703373,-0.04296062856397809), NE * labelscalefactor); dot((7.,0.),linewidth(2.pt) + dotstyle); label("$C$", (7.248939641476182,0.017494414524551142), NE * labelscalefactor); dot((2.9226598752830073,0.8078021630050068),linewidth(2.pt) + dotstyle); label("$P$", (2.43,1.0250784660000383), NE * labelscalefactor); dot((4.271575759791679,-2.5441145310506474),linewidth(2.pt) + dotstyle); label("$L$", (4.266490849108739,-3.0455611019609297), NE * labelscalefactor); dot((7.026234411480838,1.942000642419889),linewidth(2.pt) + dotstyle); label("$M$", (7.107877874269614,2.0326625174755253), NE * labelscalefactor); dot((-0.12392909019013483,1.4113920232615658),linewidth(2.pt) + dotstyle); label("$N$", (-0.5497609169440922,1.125836871147587), NE * labelscalefactor); dot((2.9226598752830073,-0.8078021630050068),linewidth(2.pt) + dotstyle); label("$D$", (2.5737496426299193,-1.3326682144526016), NE * labelscalefactor); dot((5.138560360236284,3.7164888395768223),linewidth(2.pt) + dotstyle); label("$E$", (5.15,3.806010448072383), NE * labelscalefactor); dot((-1.624597948619468,2.56030595564012),linewidth(2.pt) + dotstyle); label("$F$", (-1.5371932873900702,2.6372129483608178), NE * labelscalefactor); dot((6.683310936129654,2.794350206898316),linewidth(2.pt) + dotstyle); label("$T$", (6.765299296767949,2.8790331207149347), NE * labelscalefactor); dot((5.980167730329973,-1.6528753374785081),linewidth(2.pt) + dotstyle); label("$R$", (6.241355590000695,-1.8566119212198549), NE * labelscalefactor); dot((6.062591929471243,3.610828670951718),linewidth(2.pt) + dotstyle); label("$Q$", (6.140597184853146,3.685100361895324), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] We want $\angle{PTM}=\angle{PEM}\Leftrightarrow \angle{PTL}+\angle{MTL}=\angle{PEM}\Leftrightarrow 180^{\circ}-\angle{PDL}+\angle{LAM}=\angle{PEM}\Leftrightarrow$ $\Leftrightarrow \angle{PEM}+\angle{PDL}=180^{\circ}+\angle{LAM}\Leftrightarrow \angle{PEC}+\angle{CEM}+\angle{PDC}+\angle{CDL}=180^{\circ}+\angle{LAM}\Leftrightarrow $ $\Leftrightarrow \angle{DPE}+\angle{CEM}+\angle{CDL}=180^{\circ}+\angle{LAM}\overset{PE\perp AC,PD\perp BC}{\Leftarrow \!=\!=\!=\!=\! \Rightarrow}180^{\circ}-\angle{C}+\angle{CEM}+\angle{CDL}=180^{\circ}+\angle{LAM}$ $\Leftrightarrow \angle{CEM}+\angle{CDL}=\angle{C}+\angle{LAM}$. However,$\angle{C}+\angle{LAM}=\angle{C}+\angle{PAC}+\angle{CBP}=\angle{APB}=\angle{MPL}$,hence,the above relation is equivalent to $\angle{CEM}+\angle{CDL}=\angle{MPL} \ (1)$. However,$\angle{CEM}+\angle{QEM}=180^{\circ}-\angle{AEC}=180^{\circ}-\angle{APC}=\angle{CPL}$ and $\angle{CDL}-\angle{LDR}=\angle{CDR}=180^{\circ}-\angle{BDC}=180^{\circ}-\angle{BPC}=\angle{MPC}$,so $(1)$ is equivalent to $\angle{CPL}-\angle{QEM}+\angle{MPC}+\angle{LDR}=\angle{MPL}\Leftrightarrow \angle{LDR}=\angle{QEM}\Leftrightarrow \angle{BDL}=\angle{AEM}$. Lemma:The triangles $\triangle{BDL},\triangle{AEM}$ are similar. Proof:Firstly,observe that $\angle{LBD}=\angle{LBC}-\angle{DBC}=\angle{LAC}-\angle{CBP}=\angle{EAC}-\angle{CBM}=$ $=\angle{EAC}-\angle{CAM}=\angle{EAM}$. Hence,it suffices to show that $\frac{BL}{AM}=\frac{BD}{AE}\Leftrightarrow \frac{BL}{AM}=\frac{BP}{AP}$ which is true because $\triangle{APM}\simeq \triangle{BPL}$. From the lemma we obtain $\angle{BDL}=\angle{AEM}$,q.e.d. Thus $PETM$ is cyclic.Similarly,$FTPN$ is cyclic,an we are done.

my sol =)))

Let $X$ be the anti-Steiner point of $HP$, and let $H_A$ be the reflection of the orthocenter $H$ across $BC$. Note that $H_A D$ is the reflection of $HP$ across $BC$, so it meets the circumcircle again at $X$. But $AH_A\parallel PD$, so by Reim's theorem $PDLX$ is cyclic, and $T=X$.

Invert thru $S$ with power $\sqrt{-P_{\odot ABC}{(S)}}$. Let $O_{A},O_{B},O_{C}$ be the circumcenters of $\odot SB_{1}C_{1},\odot SC_{1}A_{1},\odot SA_{1}B_{1}$ ,we need to prove that $ABC$ and $O_{A}O_{B}O_{C}$ are perspective.$O_{A}O_{B}\perp SB$ and $O_{B}S\perp AC$ $\implies$ $O_{A}O_{B}O_{C}$ and $ABC$ are orthologic and they orthology centers are same i.e S.By Sondat's theorem (second part) we get the result.

Computational solution: We wish to show the circumcenters of $(PDL), (PEM), (PFN)$ are collinear. Remark that the circumcenter of $PDL$ is just the intersection of $BC$ with the perpendicular bisector of $PL$. So if we define $X$ to be the circumcenter of $(PMN)$ and $Y,Z$ similarly, then the circumcenters in question are just $YZ\cap BC, ZX\cap CA, XY\cap AB$, and it suffices to show by Desargues that $AX,BY,CZ$ concur. We do this by trig ceva on $\triangle XYZ$. Let $A_1,B_1,C_1$ be the midpoints of $PL,PM,PN$. We want $\displaystyle\prod_{cyc} \dfrac{\sin \angle C_1XA}{\sin \angle AXB_1}=1$. Remark that $\dfrac{\sin \angle C_1XA}{\sin \angle AXB_1} = -\dfrac{\text{dist} (A, XC_1)}{\text{dist} (A, XB_1)}$, and $\text{dist} (A, XC_1) = AP \sin \angle (AP, XC_1) -C_1P = - AP \cos \angle APC - C_1P$. Similar results hold for the other terms, so our cyclic product is just $\displaystyle\prod_{cyc} \dfrac{AP \cos \angle APC + C_1P}{AP \cos \angle APB + B_1P}$. By Law of Cosines on each term, this amounts to $\displaystyle\prod_{cyc} \dfrac{\frac{AP^2+CP^2-b^2+ 2C_1P \cdot CP}{CP}}{\frac{AP^2+BP^2-c^2 + 2B_1P \cdot BP}{BP}}$. But this is clearly $1$ because $2AP \cdot A_1P = 2BP \cdot B_1P = 2CP \cdot C_1P$ by Power of a Point, so we are done.

An old post https://artofproblemsolving.com/community/u57217h330329p1767948

[asy][asy] unitsize(2.5inches); pair A=dir(115); pair B=dir(220); pair C=dir(-40); pair H=orthocenter(A,B,C); pair P=0.25*A+0.37*B+0.38*C; pair D=2*foot(P,B,C)-P; pair HH=2*foot(H,B,C)-H; pair L=2*foot(0,A,P)-A; pair T=2*foot(0,HH,D)-HH; draw(circumcircle(A,B,C)); draw(A--B--C--cycle); draw(H--P); draw(A--HH--D); draw(D--P,dotted); draw(A--L); dot("$A$",A,dir(135)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(180)); dot("$P$",P,dir(20)); dot("$T$",T,dir(T)); dot("$H'$",HH,dir(HH)); dot("$D$",D,dir(D)); dot("$L$",L,dir(L)); [/asy][/asy] Let $T$ be the anti-Steiner point of $HP$. We claim that $T\in(PDL)$. Let $H'$ be the reflection of $H$ in $BC$. Note that $\angle PDT=\angle HPD$, which is the angle between the Steiner lines of $A$ and $T$, which is $\frac{1}{2}\widehat{AT}=\angle PLT$. Thus, $\angle PDT=\angle PLT$, as desired. Thus, all the circles concur at the anti-Steiner point of line $HP$.

If we say that the circumcenter of those three circles are collinear we are done . For prove that let Ta second intersection of BC with circle PDL . in the same way we name Tb , Tc . We know that PLTa=90 so if A' is a point on circle ABC such that ABA'=90 , we have Ta , L , A' are collinear . Now we have ratio of TaC to TaB . (LB.A'B.CTa=TaB.LC.A'C) With use of ceva's theorem we can easily find that Ta , Tb , Tc are collinear , so the circumcenters are , and the proof is finished .

I'm disappointed that no complex bash has been posted yet.

Let $T$ be the anti-steiner point of $P$, then the reflection of $H$ in any one side, $T$ and one of the points $ \{ D,E,F \}$ are collinear. Now by trivial angle chasing it is easy to see that each of the circles pass through $T$.

Let $T$ be the anti-Steiner point of $P$. We claim that $T$ is the desired concurrency point. Let $H$ be the orthocenter, $H'_A$ be the reflection of $H$ about $BC$. Then $T$ lies on $DH'_A$ since it is the reflection of $PH$ in $BC$, so $\measuredangle PDT = \measuredangle AH'_AT = \measuredangle ALT = \measuredangle PLT$, so $T$ lies on $(PDL)$. Similarly, $T$ lies on $(PEM)$ and $(PFN)$.