This problem is so cute Congrats to the proposers!

Solved with MrOreoJuice and old friend Executioner230607. Solution: Let $T \coloneqq EF \cap BC$ and denote the inverse of any point $\bullet$ as $\bullet '$ across $\omega$ Claim: $X,I,C$ are collinear, similarly $Y, I ,B$ are collinear. Proof: Note that \[\measuredangle XDE = \measuredangle XDI + \measuredangle IDE = \measuredangle IEX + \measuredangle DEI = \measuredangle DEX \]hence $\overline{XD} = \overline{XE}$ therefore $X$ lies on the perpendicular bisector of $\overline{DE}$ which is also $CI$. $\square$ Claim: $XQED$ is cyclic, similarly $YQFD$ is cyclic. Proof: By Sharky-Devil and by using the fact that $Q$ is the Miquel Point of $BCFE$ we have, \begin{align*} \angle DQE &= \angle DQC + \angle CQE \\ &= \angle A/2 + \angle BQF \\ &= \angle A/2 + \angle BTF \\ &= \angle A/2 + (\angle B - \angle BFT) \\ &= \angle A/2 + (\angle B - \angle AFE) \\ &=90^\circ - \angle C \end{align*}and \begin{align*} \angle DXE &= 2\angle DXI \\ &=2(\angle DIC - 45^\circ) \\ &=2(90^\circ - \angle C/2 - 45^\circ) \\ &= 90^\circ - \angle C \end{align*}therefore $\angle DQE = \angle DXE$. $\square$ Consider an inversion around $\omega$. $Q'$ is foot of $D$ to $\overline{EF}$. $X'$ is intersection of $CI$ and $(Q'ED)$ which is intersection of perpendicular bisector of $DE$ with semicircle with diameter $DE$ (and from our original diagram this intersection lies inside $\triangle DEF$). $A'$ is midpoint of $\overline{EF}$. $Y'$ has a similar definition as $X'$. Therefore in this inverted figure, we need to prove $A'X'Y'Q'$ is cyclic. We will now complex bash the inverted problem with a little bit of synthetic magic! Observe that \[\measuredangle X'QY' = \measuredangle X'QA' + \measuredangle A'QY' = \measuredangle X'DE + \measuredangle Y'DF = -45^\circ + (-45)^\circ = 90^\circ\] We now need to show that $\measuredangle XA'Y' = 90^\circ$. Start by computing $X'$ and $Y'$. Note that $X'$ is precisely the clock-wise rotation of $D$ across midpoint of $\overline{DE}$. Similarly, $Y'$ is counter-clockwise rotation of $D$ across midpoint of $\overline{DF}$. This yields \[x' = \left( \frac{e-d}{2}\right)i + \frac{e+d}{2} \qquad \qquad y' = \left( \frac{d-f}{2}\right)i + \frac{d+f}{2}\]To show that $A'X' \perp A'Y'$ it is enough to show that $\frac{x'-a'}{y'-a'}$ is purely imaginary. Using $a' = \frac{e+f}{2}$ This is true since \[\frac{x'-a'}{y'-a'}= \frac{\left(\frac{e-d}{2}\right)i + \frac{d-f}{2}}{\left(\frac{d-f}{2}\right)i + \frac{d-e}{2}} = i\]and thus we're done. $\blacksquare$

Just spam inversion lol Let's do inversion around $\omega$.Let $Z=(AEF)\cap (ABC)$ Then $A\to A'$=midpoint of $EF$,$Z\to Z'$=the foot from $D$ to $EF$ and since $\angle IDX=\angle IEX=45$ we have that $\angle IX'D'=\angle IX'E'=45$ so $D'X'E'=90$ so $X'$ is on the circle with diameter $DE$.Since $X'I'$ is the bisector of $D'X'E'$ we have that $X'I'$ intersects the circle with diameter $DE$ in the middle of the arc $DE$.Since $I$ and the middle of the arc are on the perpendicular bisector of $DE$ then $X'$ is the middle of the bigger arc $\widehat{DE}$ and we want to prove that $X'Y'Z'A'$ is cyclic Let's restate the problem as follows: We have a triangle $ABC$ with $D$ the foot from $A,M$ the midpoint of $BC$,$X$ the midpoint of arc $\widehat{ADC}$,$Y$ the midpoint of arc $\widehat{ADB}$ and we want to prove $XDMY$ cyclic.For this we will do inversion at $D$ with power $1$ The problem becomes: Triangle $A'B'C',D$ the foot from $A'$,since $\angle ADX=45$ we have $\angle A'DX'=45$ so $DX'$ is the exterior bisector of $\angle ADC$ with $X'\in A'C'$,similary $Y'$ and we want $X',Y',M'$ collinear For this we do menelaus: We want $\frac{A'Y'}{Y'B'}\cdot \frac{B'M'}{M'C'}\cdot \frac{C'X'}{X'A'}=1$. By exterior angle theorem $\frac{A'X'}{X'C'}=\frac{A'D}{DC'},\frac{A'Y'}{Y'B'}=\frac{A'D}{DB'}$ and so we want to prove that $\frac{M'B'}{M'C'}=\frac{DB'}{DC'}$,but $M'B'=DB'+DM'=\frac{1}{DB}+\frac{1}{DM},M'C'=\frac{1}{DM}-\frac{1}{DC}$,$\frac{DB'}{DC'}=\frac{DC}{DB}$ and plugging in gives the result.

Rijul saini wrote: Let $ABC$ be an acute angled triangle with $AC>AB$ and incircle $\omega$. Let $\omega$ touch the sides $BC, CA,$ and $AB$ at $D, E,$ and $F$ respectively. Let $X$ and $Y$ be points outside $\triangle ABC$ satisfying \[\angle BDX = \angle XEA = \angle YDC = \angle AFY = 45^{\circ}.\]Prove that the circumcircles of $\triangle AXY, \triangle AEF$ and $\triangle ABC$ meet at a point $Z\ne A$. Proposed by Atul Shatavart Nadig and Shantanu Nene $\textbf{Claim 1:}$ Let $L$ be foot of altitude from $D$ to $EF$; $Z$ be the second intersection of circles $(AEF)$, $(ABC)$ $\hspace{1.5cm}$Then $Z$, $L$, $I$ are colinear Proof : $\, \,$Cuz $\angle FLD = 90^\circ$ and $L(FD, BC) = -1$ so $LD$ is the internal bisector of $\angle BLC$ (Cuz $D$ lie on segment $BC$) $\, \,$Which means $\angle FLB = \angle ELC$ $\, \,$Combine with $\angle AFE = \angle AEF$, we see that $\triangle LFB \sim \triangle LEC$ $\, \,$In the order hand, cuz $Z = (AEF) \cap (ABC)$ ($Z$ \neq $A$) so $Z$ is center of spiral similarity carries $(AEF)$ to $(ABC)$, $\, \,$which means $\triangle ZFB \sim^+ \triangle ZEC$ $\, \,$Thus, $\frac{LF}{LE} = \frac{FB}{EC} = \frac{ZF}{ZE}$. $\, \,$So we have that $ZL$ is the internal bisector of $\angle FZE$ and $Z$, $L$, $I$ are colinear (Cuz $L$ always lie on segment $EF$) $\textbf{Claim 2:}$ $X$, $Z$, $E$, $D$ are concyclic; $Y$, $Z$, $F$, $D$ are concyclic Proof : $\, \,$Cuz $Z$ is center of spiral similarity carries $EF$ to $BC$ and $\frac{LF}{LE} = \frac{FB}{EC} = \frac{DB}{DC}$ $\, \,$So $\triangle ZLD \sim^+ \triangle ZFB$. Lead to $\angle DZL = \angle BZF$ $\, \,$Therefore, $\angle FZD = \angle FZI - \angle DZL = \frac12 \angle BAC - \angle BZF = \frac12 \angle BAC + \angle FZA - \angle BZA $ $\hspace{3cm}= \frac12 \angle BAC + \angle ACB - \angle AEF = \angle BAC + \angle ACB - 90^\circ = 90^\circ - \angle ABC$ $\, \,$In the order hand, we see that $\angle FDY = \angle FDI + \angle IDY = \frac12 \angle ABC + 45^\circ, \angle DFR = \frac12 \angle ABC + 45^\circ$ $\, \,$so $\angle DYF = 90^\circ - \angle ABC$ $\, \,$Thus, $\angle DZF = 90^\circ - \angle ABC = \angle DYF$. Lead to $Y$, $Z$, $F$, $D$ are concyclic. Similar, $X$, $Z$, $E$, $D$ are concyclic $\textbf{Claim 3:}$ Let $X'$, $Y'$ be points inside $\angle FDE$ satisfying $\triangle DX'E$, $\triangle DY'F$ are right isosceles triangles at $X$, $Y$ $\hspace{1.5cm}$Then $A'$, $X'$, $Y'$, $L$ are concyclic, with $A'$ is the midpoint of segment $EF$ Proof : $\, \,$Let $U$, $V$ be the reflections of $E$, $F$ across $X'$, $Y'$, respectively $\, \,$We easy to see that $\triangle EDU$, $\triangle FDV$ are right isosceles triangles at $D$ $\, \,$Lead to $D$ is a center of rotation with $90^\circ$ carries $UF$ to $EV$ $\, \,$Therefore $EV \perp FU$ and $EV = FU$. Which means $\triangle X'A'Y'$ is a right isosceles triangle (Cuz $\overrightarrow{A'X'} = \frac12 \overrightarrow{EV}; \overrightarrow{A'Y'} = \frac12 \overrightarrow{FU}$) $\, \,$But we ez to see that $\angle X'LY' = \angle X'LD + \angle Y'LD = \angle X'ED + \angle Y'FD = 45^\circ + 45^\circ = 90^\circ$ $\, \,$Thus, $X'$, $Y'$, $A'$, $L$ are concyclic on a circle with diameter $X'Y'$ $\textbf{From these claim we see that :}$ $\, \,$Let $f$ be the inversion about the incircle of $\triangle ABC$. $\, \,$Cuz $\angle AZI = \angle IA'L = 90^\circ$ so $ID^2 = \overline{IA'}.\overline{IA} = \overline{IL}.\overline{IZ}$ $\, \,$Therefore, $f$ : $A \leftrightarrow A'$, $Z \leftrightarrow L$, $(ZFD) \leftrightarrow (DF)$, $(ZED) \leftrightarrow (DE)$. $\, \,$Combine with $X$ and $X'$ lie on perpendicular bisector of segment $DF$, $Y$ and $Y'$ lie on perpendicular bisector of segment $DE$ $\, \,$we see that $f$ : $X \leftrightarrow X'$, $Y \leftrightarrow Y'$. $\, \,$But in Claim 3 we proved that $X'$, $Y'$, $L$, $A'$ are concyclic $\, \,$Thus, $A$, $Z$, $X$, $Y$ are concyclic

complex is cool, but i am too lazy to type

Same configuration and series of lemmas used in USA TSTST 2020 P2 Edit: Well it wasn't nearly as simple as I thought it would be after inversion coz u need to angle chase a bunch afterwards.

L13832 wrote: Same configuration and series of lemmas used in USA TSTST 2020 P2 Also similar to Kazakhstan National Olympiad 2024 (9 grade), P5

Invert with respect to the incircle and restate in terms of $\Delta DEF$. Inverted Problem wrote: In $\Delta ABC$, $X$ and $Y$ are midpoints of arcs $AB$ and $AC$ of the circles $(AB)$ and $(AC)$ lying on the same side of the side as the opposite vertex. Let $D$ be the foot of altitude from $A$ onto $BC$ and $M$ be the midpoint of $BC$. Prove that $X,Y,M,D$ are concyclic. Note that by simple angle chasing, $\angle XDY=90$° so it suffices to show that $\angle XMY=90°$°. Note that upon fixing $AB,AC,BC$; $AY,CY,BX,AX$ are known in terms of $a,b,c$. Since angle $BAY$ can be worked out, by cosine formula we can calculate $BY$ and similarly $CX$. By apollonius theorem in $\Delta BYC$, we can calculate $YM$ and similarly $XM$. Lastly, one can also find out $XY$ using cosine fomula in $\Delta AXY$ so we are done by converse of pythagoras in $\Delta XMY$.

Invert around the incenter. Note that $(AEF)\cap (ABC)$ is $A-$sharky devil point which swaps with the foot of the altitude from $D$ to $EF$ under this inversion. New Problem Statement: $ABC$ is a triangle with altitude $AD$ where $D$ lies on $BC$. Let $X,Y$ be outside $\triangle ABC$ such that $\measuredangle AXB=90=\measuredangle CYA$ and $AX=AB,YA=YC$. If $M$ is the midpoint of $BC$, then prove that $D,M,X,Y$ are concyclic. Invert at $D$ with radius $\sqrt{-DB.DC}$. New Problem Statement: $ABC$ is a triangle with altitudes $AD,BE,CF$ and $D,E,F$ lie on $BC,CA,AB$ respectively. $X,Y$ lie on $CF,BE$ such that $DX,DY$ bisect $\measuredangle ADC,\measuredangle BDA$ respectively. Prove that $BC,EF,XY$ are concurrent. Let $H$ be the orthocenter of $\triangle ABC$. Set $EF\cap BC=Q$. \[\frac{QB}{QC}.\frac{XC}{XH}.\frac{YH}{YB}=\frac{QB}{QC}.\frac{DC}{DH}.\frac{DH}{DB}=\frac{QB}{QC}.\frac{DC}{DB}=1\]As desired.$\blacksquare$