hatchguy wrote:

$RSM$ 's post made me realize there's a really easy proof by inversion.

Firstly note that $\frac{A_1B}{A_1C} \cdot \frac{C_1A}{C_1B} \cdot \frac{CB_1}{AC_1} = 1$ is equivalent to trigonometric ceva.

From this we have $\frac{A_2B}{A_2C} \cdot \frac{C_2A}{C_2B} \cdot \frac{CB_2}{AC_2} = 1$ (*).

As $RSM$ said, $A_2$ is in the circumcircle of $BHC$ . Similarly for the others. We perform an inversion centered at $H$ . We have, for example, $B', C', A_2'$ are collinear. We need to show that $A_2', B_2', C_2'$ are collinear.

Applying menelaus for this points in triangle $A'B'C'$ we are left to prove $\frac{A_2B}{A_2C} \cdot \frac{C_2A}{C_2B} \cdot \frac{CB_2}{AC_2} = 1$ which is what we have in (*) and we are done.

Here's another solution which is essentially the above one, but fleshed out more.

$[asy] unitsize(2.5inches); pair A=dir(115); pair B=dir(220); pair C=dir(-40); pair H=orthocenter(A,B,C); pair P=0.27*A+0.35*B+0.38*C; pair D=2*foot(H,B,C)-H; pair A1=2*foot(0,A,P)-A; pair X=foot(A,B,C); pair Y=foot(B,A,C); pair Z=foot(C,A,B); pair A2=2*foot(A1,B,C)-A1; pair A3=extension(Y,Z,A2,H); pair U=extension(A,H,Y,Z); draw(circumcircle(A,B,C)); draw(A--D); draw(A--B--C--cycle); draw(C--Z); draw(B--Y); draw(A--A1); draw(A1--A2); draw(A2--A3); draw(Y--A3); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(125)); dot("$P$",P,dir(0)); dot("$D$",D,dir(D)); dot("$A_1$",A1,dir(A1)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); dot("$Z$",Z,dir(140)); dot("$U$",U,dir(145)); dot("$A_2$",A2,dir(45)); dot("$A_3$",A3,dir(A3)); [/asy]$

We'll relabel $X$ in the problem to $P$ . Let $H$ be the orthocenter of $\triangle ABC$ with orthic triangle $XYZ$ . Let $D=AH\cap(ABC)$ , and define $E$ , $F$ similarly. Also let $U=AH\cap YZ$ , and define $V$ , $W$ similarly. Finally, let $A_3=HA_2\cap YZ$ , and define $B_3$ and $C_3$ similarly.

We claim that $H$ is the desired fixed point. To see this, we'll invert about $H$ with (negative) power $HA\cdot HD$ . Note that $H$ and $D$ are reflections over $BC$ , so $A_2\in(BHC)$ . The image of $(BHC)$ is $YZ$ , so the image of $A_2$ is $A_3$ . Thus, it suffices to show that $A_3$ , $B_3$ , $C_3$ collinear.

By Menalaus's theorem, this is equivalent to showing that $\prod_{\mathrm{cyc}}YA_3/A_3Z=-1$ . We already know that $\prod_{\mathrm{cyc}}YU/UZ=1$ by Ceva, so it suffices to show that $\prod_{\mathrm{cyc}}(YZ;UA_3)=-1$ . We now compute
$(YZ;UA_3)\stackrel{H}{=}(B,C;X,HA_2\cap BC)=(B,C;X,DA_1\cap BC)\stackrel{D}{=}(BC;A_1A).$ It's easy to see that the concurrence of $AA_1$ , $BB_1$ , $CC_1$ is equivalent to $\prod_{\mathrm{cyc}}(BC;A_1A)=-1$ (say using trig Ceva), so we're done.