Overkill hint

There is a solution with Moving points

Almost gave into my temptation to trig ceva bash . The problem is actually trivial with Jacobi's. Solution Let incircle $\Omega$ touch $\overline{AB}$, $\overline{BC}$, and $\overline{CA}$ at $C''$, $A''$, and $B''$, respectively. Note that since $\overline{IA''}\perp\overline{BC}$ and $\overline{IA'}\perp\overline{BC}$, $A''$, $A'$, and $I$ are collinear. Note that since $\overline{IB''}\perp\overline{AC}$ and $\overline{IB'}\perp\overline{AC}$, $B''$, $B'$, and $I$ are collinear. Note that since $\overline{IC''}\perp\overline{AB}$ and $\overline{IC'}\perp\overline{AB}$, $C''$, $C'$, and $I$ are collinear. Thus we see that since $A'A''=B'B''=C'C''$ and $B''C=A''C$, $C''B=A''B$, and $B''A=C''A$, $\angle A'CA'' = \angle B'CB''$, $\angle A'BA'' = \angle C'BC''$, and $\angle C'AC'' = \angle B'AB''$. Thus lines $AA', BB', CC'$ concur by Jacobi's theorem. $\blacksquare$

Normal Problem wrote: Let $ABC$ be a triangle with incenter $I$ and intouch triangle $DEF.$ Choose $X,Y,Z$ on rays $ID,IE,IF$ respectively with $IX=IY=IZ.$ Show that lines $AX,BY,CZ$ concur. Dr_Vex wrote:

Overkill hint

Then Invert Problem wrote: Let $DEF$ be a triangle with circumcenter $I$.The points $A,B,C$ is a midpoints of the sides $FE,FD,ED$.Choose $X,Y,Z$ on rays $ID,IE,IF$ respectively with $IX=IY=IZ.$ Show that circles $\odot(IAX),\odot(IBY),\odot(ICZ)$ concur.($\odot(IAX)\cap \odot(IBY)\cap \odot(ICZ)=N,N\neq I$ Can Dr_Vex tell you the solution to this Invert problem?