easy by calculation(sorry the last line should be sin(π+C)/2 but I'm too lazy to reupload picture)

This can't be intentional but this and 2024 EGMO/2 are extremely similar. Let $H=AM\cap GZ$ and $S$ be the sharkydevil point. It suffices to show that $GMDH$ is cyclic since then $\angle ZMH=\angle MDC$ which is well known to be $\angle SAM$ by inversion at $M$. Let $K$ be $XY\cap BC$; by EGMO 2 we have that $K$ is the circumcenter of right triangle $DGS$ where $D$ is the intouch point. Claim: $SGZK$ is cylic Proof.We use $XY\parallel AT\parallel EF$. Note that \[\measuredangle KZS = \measuredangle ITS = \measuredangle IDS = \measuredangle DSI = \measuredangle DGS= \measuredangle KGS\]with the second to last equality coming from $\angle GSD = \angle IDG = 90^{\circ}$. $\blacksquare$ To finish, \[\measuredangle HGD = \measuredangle ZGK = \measuredangle ZSK = \measuredangle (SA, KS) = \measuredangle AMS = \measuredangle HMD\]with $\measuredangle (SA, KS) = \measuredangle AMS$ holding from $KS$ is tangent to $(ABC)$ by EGMO 2.

GrantStar wrote: This can't be intentional but this and 2024 EGMO/2 are extremely similar. Let $H=AM\cap GZ$ and $S$ be the sharkydevil point. It suffices to show that $GMDH$ is cyclic since then $\angle ZMH=\angle MDC$ which is well known to be $\angle SAM$ by inversion at $M$. Let $K$ be $XY\cap BC$; by EGMO 2 we have that $K$ is the circumcenter of right triangle $DGS$ where $D$ is the intouch point. Claim: $SGZK$ is cylic Proof.We use $XY\parallel AT\parallel EF$. Note that \[\measuredangle KZS = \measuredangle ITS = \measuredangle IDS = \measuredangle DSI = \measuredangle DGS= \measuredangle KGS\]with the second to last equality coming from $\angle GSD = \angle IDG = 90^{\circ}$. $\blacksquare$ To finish, \[\measuredangle HGD = \measuredangle ZGK = \measuredangle ZSK = \measuredangle (SA, KS) = \measuredangle AMS = \measuredangle HMD\]with $\measuredangle (SA, KS) = \measuredangle AMS$ holding from $KS$ is tangent to $(ABC)$ by EGMO 2. This is indeed not intentional. I personally haven't checked EGMO's problems except for Problem 6.

Nice problem... Let sharkydevil point be $S = (AEF) \cap (ABC)$ and $K = XY \cap BC$. Applying radical thm on $(AEF)$, $(BCXY)$ and $(ABC)$ we have $K$ lie on $AS$. Define $D$ be incircle touch point and $J = AT \cap EF$. $A' = AI \cap (ABC)$. Note from $KD^2=KX.KY=KB.KC$ and it's well known that $(G,K;B,C)=-1$ $\implies$ $K$ is midpoint of $DG$. It's well known that $SD$ bisects $\measuredangle BSC$ hence by Apollonian circle property we have $\measuredangle GSD = 90$. $$\measuredangle KZS = 90 - \measuredangle SAI = 90 - \measuredangle SEI = 90 - \measuredangle SCA' = 90 - \measuredangle SDG = \measuredangle DGS$$ hence $G,K,S,Z$ cyclic Note we want to prove $2 \measuredangle SAI = \measuredangle GZA$, but as $AI \perp EF$ we just have to prove $\measuredangle ZGJ = \measuredangle GJZ$. I didn't find any other solu then angle bash $$\measuredangle ZGJ = \measuredangle TGJ + \measuredangle ZGD = \measuredangle TGJ + \measuredangle TSK =\measuredangle TGJ + \measuredangle DKS - \measuredangle DTS = $$$$\measuredangle DKS - \measuredangle GJT = \measuredangle DKS + \measuredangle JAI - 90 = 90 - 2 \measuredangle SDK + \measuredangle JAI =$$$$ 90 - \measuredangle A - 2\measuredangle SAF + \measuredangle JAI = 90 - \measuredangle SAI = \measuredangle FJA = \measuredangle GJZ$$$\blacksquare$

Let $S$ be the $A$-Sharkydevil point. It is well-known that $S$ lies on $AT$. Let $XY \cap BC = M$, then since $MD^2 = MX \cdot MY = MB \cdot MC$ and $(G,D;B,C)=-1$, we know $M$ is the midpoint of $GD$. Let $AI \cap BC = L$ and $J$, $K$, $H$ be the midpoints of $AT$, $AL$, $AI$ respectively, then $JH \perp AI$ and it suffices to show that $JI \parallel GZ$. Let $N$ be the projection of $A$ on $ID$ and let $O = ID \cap EF$, then it is well-known that $N$ and $I$ are inverses w.r.t. the incircle. Now if $\infty$ is the point at infinity along $BC$, then \[ Z(M,T;G,\infty) = \frac {MG}{TG} = \frac {DG}{2TG} = \frac{DO}{2IO} = \frac{ND}{2ID} = \frac{AL}{2IL} = \frac{AK}{2HK} = \frac{HI}{AI} \cdot \frac{AK}{HK} = J(H,A;I,K) \]and as $ZM \parallel JH, ZT \parallel JA, Z\infty \parallel JK$, we conclude that $ZG \parallel JI$, as desired.

Li4 wrote: Let $I$ be the incenter of triangle $ABC$, and let $\omega$ be its incircle. Let $E$ and $F$ be the points of tangency of $\omega$ with $CA$ and $AB$, respectively. Let $X$ and $Y$ be the intersections of the circumcircle of $BIC$ and $\omega$. Take a point $T$ on $BC$ such that $\angle AIT$ is a right angle. Let $G$ be the intersection of $EF$ and $BC$, and let $Z$ be the intersection of $XY$ and $AT$. Prove that $AZ$, $ZG$, and $AI$ form an isosceles triangle. Proposed by Li4 and usjl. $\textbf{Claim 1:}$ Let $D$ is the point of tangentcy of $\omega$ with $BC$; $N$ is the foot of altitude from $D$ to $EF$ $\hspace{2.1cm}A'$, $R$ are midpoints of segments $EF$, $XY$ respectively; $P$, $I'$ are refelections of $N$, $I$ through $A'$, $XY$ respectively $\hspace{1.5cm}$Then $PI' \| ID$ Proof : $\, \,$Let $B'$, $C'$ in order are midpoints of segments $DF$, $DE$ and $I_a$ is the A-excenter of $\triangle ABC$ $\, \,$Cuz $ IX^2 = IY^2 = \overline{IR}. \overline{II_a} = \overline{IC'}. \overline{IC} =\overline{IB}. \overline{IB'} = ID^2$ $\, \,$So combine with $B$, $C$, $X$, $Y$, $I$ are concyclic, invert about $\omega$, we see that $X$, $Y$, $B'$, $C'$ are colinear $\, \,$Which means $XY$ is the D-midline of $\triangle DEF$ $\, \,$In the order hand, $IR$ is the perpendicular bisector of segment $EF$, $DN$ is a altitude of $\triangle DEF$ $\, \,$Thus, $R$ is midpoint of segment $DP$. Lead to $IPI'D$ is a parallelogram, $PI' \| DI$ $\textbf{Claim 2:}$ Let $V$ is the intersection of lines $AN$, $DI_a$; $U$ is the intersection of lines $AP$, $ID$ $\hspace{1.5cm}$Then $V$, $A'$, $U$ are colinear Proof : $\, \,$Let $I_b$, $I_c$ are B-excenter, C-excenter of $\triangle ABC$ respectively and $H$ is the orthocenter of $\triangle DEF$ $\, \,$Cuz $EF \| I_bI_c$, $FD \| I_cI_a$, $DE \| I_aI_b$ and $A$, $N$ are foots of altitudes from $I_a$, $D$ to $I_bI_c$, $EF$ $\, \,$So $V$ is a homothety centered that takes $\triangle I_aI_bI_c $ $\cup$ {$A$, $I$} to $\triangle DEF$ $\cup$ {$N$, $H$} $\, \,$(Cuz $I$ is the orthocenter of $\triangle I_aI_bI_c$) $\, \,$Thus, combine with Claim 1, we have $\frac{\overline{VN}}{\overline{VA}} = \frac{\overline{HN}}{\overline{IA}} = \frac{\overline{I'I}}{\overline{IA}} = - \frac{\overline{UP}}{\overline{UA}}$ $\, \,$Therefore, use Menelaus theorem for $\triangle ANP$, we see that $V$, $U$, $A'$ are colinear $\textbf{Claim 3:}$ Let $M$ is the intersection of lines $IaN$, $AD$. Then $II_a$ is the exterior bisector of $\angle NIM$ Proof: $\, \,$Cuz $A' = II_a \cap PN$, $U = AP \cap ID$, $V = AN \cap I_aD$ and $V$, $U$, $A'$ are colinear (From Claim 2) $\, \,$So use Dersagues theorem for triangles $IDN$, $PAI_a$, we see that lines $AD$, $NI_a$, $IP$ concur at $M$ $\, \,$But $II_a$ is the internal bisector of $\angle PIN$, therefore $II_a$ is the exterior bisector of $\angle NIM$ $\textbf{From these claim, we have :}$ $\, \,$Cuz $IT \perp DN$ and $BC$ is the polar of $D$ with respect to $\omega$ $\, \,$So follow La Hire theorem, we see that : $DN$ is the polar of $T$ with respect to $\omega$ $\, \,$Combine with $EF$ is the polar of $A$ with respect to $\omega$, we have $AT$ is the polar of $N$ with respect to $\omega$ $\, \,$In the order hand, $\angle IaXI = \angle IaYI = 90^\circ$ so$XY$ is the polar of $I_a$ with respect to $\omega$, $Z = AT \cap XY$ $\, \,$Therefore, $NI_a$ is the polar of $Z$ with respect to $\omega$ $\, \,$Thus $AZ$, $ZG$, and $AI$ form an isosceles triangle equivalent with $II_a$ is a bisector of $\angle MIN$. Which's proved in Claim 3, done

Let $AI\cap ZG=P, \ AS\cap EF=H$, $S$ be $A-$sharky devil point and $V\in AS$ such that $VD\parallel EF$. Under the inversion around the incircle, we can see that $XY$ is the mid-base of $EF$ on $\triangle DEF$. We conlude that $HZ=ZV$. \[\measuredangle SVD=\measuredangle AHE=\measuredangle SIA=\measuredangle SDI=\measuredangle SGD\]Thus, $G,V,D,S$ are concyclic which implies $\measuredangle GVD=90$ and $\measuredangle HGV=90$. Since $HZ=ZV$, we get $ZV=ZH=ZG$. \[\measuredangle ZPA=\measuredangle GZA-\measuredangle PAZ=180-2\measuredangle ZHG-\measuredangle PAZ=180-\measuredangle ZHG-(90-\measuredangle ZHG)=90-\measuredangle ZHG=\measuredangle PAZ\]So $ZA=ZP$ as desired.$\blacksquare$