The problem is mainly: $\Delta ABC$, altitududes $AD,BE,CF$, orthocenter $H$. Let $X$ be the humpty point wrt $A$, let $\left\{X,T\right\}=CX \cap (AEF)$. Prove $\overline{D,F,T}$ We can use inversion with center $C$, power $k=CE.CA$ to prove it

Sharygin 2019 Final 10 P5 wrote: Let $AA_1, BB_1, CC_1$ be the altitudes of triangle $ABC$, and $A0, C0$ be the common points of the circumcircle of triangle $A_1BC_1$ with the lines $A_1B_1$ and $C_1B_1$ respectively. Prove that $AA_0$ and $CC_0$ meet on the median of ABC or are parallel to it. We claim that $AA_0\cap CC_0$ is the $A-\text{HM Point}$ of $\triangle ABC$. Now we Invert around $B$ with radius $\sqrt{BH\cdot BB_1}$ where $H$ is the orthocenter of $\triangle ABC$. The Problem becomes Equivalent to this. Inverted Problem wrote: $BC_1'A_1'$ is a triangle and $H',A',C'$ are the foot of altitudes from $B,A_1',C_1'$ respectively and let $B_1'$ be the orthocenter $\triangle BC_1'A_1'$ respectively. Let $\odot(BB_1'C_1')\cap A_1'C_1'=C_0'$ and $\odot(BB_1'A_1')\cap C_1'A_1'=A_0'$. Then prove that $\odot(BC'C_0')\cap \odot(BA'A_0')$ is the midpoint of $A_1'C_1'$ Here's the diagram of the Inverted Image. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); 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 = -11.37, xmax = 11.37, ymin = -6.65, ymax = 6.65; /* image dimensions */ pen ffcctt = rgb(1,0.8,0.2); /* draw figures */ draw((-0.99,2.59)--(-2.35,-1.97), linewidth(1)); draw((-2.35,-1.97)--(3.97,-1.87), linewidth(1)); draw((3.97,-1.87)--(-0.99,2.59), linewidth(1)); draw((-0.99,2.59)--(-1.0100190226369379,-1.9487977693455212), linewidth(1)); draw((-2.35,-1.97)--(0.1773053859915672,1.5403665279188727), linewidth(1)); draw((3.97,-1.87)--(-1.716512153759186,0.15404748445449412), linewidth(1)); draw(circle((-4.866504228075911,1.2633433662682538), 4.0972311204171605), linewidth(1) + red); draw(circle((2.273961277046097,1.231849312589382), 3.5352533866429328), linewidth(1) + red); draw((3.97,-1.87)--(-7.29832236335084,-2.034165301053613), linewidth(1)); draw(circle((-3.187261519541944,-1.027492631598216), 4.232518295876246), linewidth(1.2) + linetype("4 4") + green); draw(circle((0.8636545831556175,0.7108398611641138), 2.6395602173550676), linewidth(1.2) + linetype("4 4") + ffcctt); /* dots and labels */ dot((-0.99,2.59),dotstyle); label("$B$", (-1.09,3.05), NE * labelscalefactor); dot((-2.35,-1.97),dotstyle); label("$C_1'$", (-2.51,-2.41), NE * labelscalefactor); dot((3.97,-1.87),dotstyle); label("$A_1'$", (4.13,-2.09), NE * labelscalefactor); dot((-1.0100190226369379,-1.9487977693455212),dotstyle); label("$D$", (-1.05,-2.37), NE * labelscalefactor); dot((0.1773053859915672,1.5403665279188727),dotstyle); label("$C'$", (0.25,1.75), NE * labelscalefactor); dot((-1.716512153759186,0.15404748445449412),dotstyle); label("$A'$", (-2.17,0.33), NE * labelscalefactor); dot((-1.0018534158666539,-0.09745674903347251),dotstyle); label("$B_1'$", (-0.71,-0.03), NE * labelscalefactor); dot((-7.29832236335084,-2.034165301053613),dotstyle); label("$C_0'$", (-7.87,-2.37), NE * labelscalefactor); dot((0.6769068029876822,-1.9221059050160179),dotstyle); label("$A_0'$", (0.19,-2.31), NE * labelscalefactor); dot((0.9484918448234196,-1.9273566402504871),dotstyle); label("$M$", (1.17,-2.25), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy] Note that $C_0'$ is just the reflection of $A_1'$ over $H'$ because $\angle BA_1'C_1'=180^\circ-\angle BB_1'C_1'=\angle BC_0'H'$. So, we get that $\angle C_0'BC'=180^\circ-2\angle BA_1'C_1'$. Also if $M$ is the midpoint of $A_1'C_1'$ then $\angle C'MC_0'=2\angle BA_1'C_1'\implies \odot(BC_0'C')\text{ passes through the midpoint of} A_1'C_1'$. Similarly we get that $\odot(BA_0'A')\text{passes through the midpoint of} A_1'C_1'$. So, Inverting back we get that $AA_0\cap CC_0\in A-\text{HM Point}$ of $\triangle ABC\in A-\text{Median}$.

Let $D=\overline{BM}\cap (A_1B_1C_1)$, let $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of $\overline{AC}$. Note that $\overline{MA_1}$ and $\overline{MC_1}$ and are tangent to $(A_1B_1C_1H)$. Hence, Pascal's theorem on $A_0DBHA_1A_1$ yields that $D$ lies on $\overline{AA_0}$. Similarly, $D$ lies on $\overline{CC_0}$. $\blacksquare$