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A single lemma kills the problem

Let $M$ and $N$ be the midpoints of arc $BC$ not containing $A$, and arc $BAC$ respectively. Note that $H$ is the $A$-Sharky Devil point of $\Delta ABC$. Recall that $H,D,M$ are collinear. Also note that $(B,C;D,K)=-1=(B,C;M,N) \stackrel{H}{=} (B,C;D,NH \cap BC)$, so $K,N,H$ are collinear. It follows that $\measuredangle KHD = \measuredangle NHM = 90^{\circ} = \measuredangle KGD$, so $H,G,D,K$ are concyclic.

A，E，I，F，H共圆（同一法）（1） 且熟知D，K调和分割BC 由（1）我们可知ΔHEC~ΔHFB 故HB/HC=BF/CE=BD/CD 所以HD平分角BHC 由调和点列性质角KHD为直角 故角KHD=角KGD 即K，H，G，D共圆。#

parmenides51 wrote: As shown in the figure, in $\vartriangle ABC$, $AB>AC$, the inscribed circle $I$ is tangent to the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively, and the straight lines $BC$ and $EF$ intersect at point $K$, $DG \perp EF$ at point $G$, ray $IG$ intersects the circumscribed circle of $\vartriangle ABC$ at point $H$. Prove that points $H$, $G$, $D$, $K$ lie on a circle. Claim1: Notice that $H$ is the sharky-devil point. Let \(M_D\) be the midpoint of \(EF\). We show that \(\measuredangle AHI = \pi/2\). To do so, invert around the incircle \((DEF)\). As this swaps \((ABC)\) with the nine-point circle of \(DEF\), the image of \(H\) is \(G\), which is the foot of the perpendicular from \(D\) to \(EF\). Hence , \(\measuredangle AHI = - \measuredangle IM_D P = \pi/2\),$H$=$(AEF) \cap (ABC)$ Claim2 : $HD$ passes through the midpoint of arc $BC$ (not containing $A$). Since $H$ is the sharky-devil point, we have that $ \triangle HEC \sim \triangle HFB$,so $\frac{HB}{HC}=\frac{FB}{EC}=\frac{BD}{CD}$,which means $HD$ bisects $\angle BHC$. Claim3 : $(B,C;D,K)=-1$ Notice that $AD$,$BE$,$CF$ concur at $Ge$(Gergonne Point),and lines $BC$ and $EF$ intersect at point $K$,so $(B,C;D,K)=-1$ At last , since $(B,C;D,K)=-1$ and $HD$ bisects $\angle BHC$,so \(\measuredangle KHD= \pi/2\).

Invert at $D$ swapping $B,C$. Let the midpoint of arc $BC$ not containing $A$ be $M$. It is well known $D,H,M$ collinear by Sharky-devil, thus $H$ is sent to $M$ as the circumcircle is fixed. Now note that $AD,BE,CF$ concur at $X_7$ hence by complete quad $(B,C;D,K)=-1$. This cross ratio is preserved under inversion hence $(B,C;K',P_\infty)=-1$ so $K'$ is the midpoint of $BC$. Hence, it is trivial to see that $\angle DK'H'=90^\circ$, or $\angle DHK=90^\circ=\angle DGK$. Hence indeed $H,G,D,K$ lie on a circle.

Let $M$ and $N$ be the midpoints of arcs $BC$ and $BAC$. It's well known that (a) $H$ is the $A$-Sharkydevil point in $\triangle{ABC}$; (b) $H$, $D$ and $M$ are collinear; and (c) $K$, $H$ and $N$ are collinear. Now $\measuredangle DHK = \measuredangle MHN = 90^\circ = \measuredangle DGK$ as desired.