The angle condition $\angle BDH+\angle ABC=180^\circ$ means that $(BDH)$ is tangent to $AB$. Similarly, $(CEH)$ is tangent to $AC$. Since $M$ is the midpoint of $BP$, this means that $M$ is of equal powers to $(BDH)$ and the incircle of $\triangle ABC$, and $M$ lies on the radical axis of these two circles. Hence, $MD$ is the radical axis of these two circles. Similarly, $NE$ is the radical axis of $(CEH)$ and the incircle. [asy][asy] size(11cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair H = foot(A,B,C); pair G = (A+B+C)/3; pair I = incenter(A,B,C); pair P = foot(I,A,B); pair Q = foot(I,A,C); pair M = (B+P)/2; pair N = (C+Q)/2; pair X = intersectionpoints(G--H+10*(H-G),unitcircle)[0]; pair D = intersectionpoints(circumcircle(B,H,X),incircle(A,B,C))[1]; pair E = intersectionpoints(circumcircle(C,H,X),incircle(A,B,C))[0]; draw(A--B--C--A--cycle); draw(A--H); draw(circumcircle(B,H,X)); draw(circumcircle(C,H,X)); draw(incircle(A,B,C)); pair T = extension(M,D,N,E); draw(M--T--N); draw(H--G); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$H$",H,dir(H)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); dot("$M$",M,dir(M)); dot("$N$",N,dir(N)); dot("$G$",G,dir(G)); [/asy][/asy] It remains to show that $HG$ is the radical axis of $(BDH)$ and $(CEH)$, from which the conclusion follows by radical axis theorem. Consider the other intersection point $X$ of $(BDH)$ and $(CEH)$. Note that \[\angle BXH+\angle CXH=\angle B+\angle C=180^\circ-\angle A.\]Therefore, $ABXC$ is cyclic. We wish to show that $GH$ intersects $(ABC)$ at $X$ as well. Equivalently, if $GH$ intersects minor arc $\widehat{BC}$ at $X'$, we shall show that $(X'BH)$ is tangent to $AB$. [asy][asy] size(9cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair H = foot(A,B,C); pair G = (A+B+C)/3; pair I = incenter(A,B,C); pair P = foot(I,A,B); pair Q = foot(I,A,C); pair M = (B+P)/2; pair N = (C+Q)/2; pair X = intersectionpoints(G--H+10*(H-G),unitcircle)[0]; pair D = intersectionpoints(circumcircle(B,H,X),incircle(A,B,C))[1]; pair E = intersectionpoints(circumcircle(C,H,X),incircle(A,B,C))[0]; draw(A--B--C--A--cycle); draw(A--H); pair Y = intersectionpoints(G--G+10*(G-H),unitcircle)[0]; draw(unitcircle); draw(circumcircle(B,X,H)); draw(A--Y--X); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$H$",H,dir(H)); dot("$G$",G,dir(G)); dot("$X$",X,dir(X)); dot("$Y$",Y,dir(Y)); [/asy][/asy] Let $Y$ be the intersection of $GH$ with major arc $\widehat{BAC}$. It is well-known that $AYCB$ is a trapezium. Therefore, \[\angle YX'B=\angle YCB=\angle ABC\]which shows our desired conclusion.

Let $\odot(BDH)\cap \odot(CEH)=L$,$T\in \odot(ABC)$ satisfy $AT//BC$ Note that $\angle BLH+\angle CLH=\angle ABC+\angle ACB=180^{\circ}-\angle BAC$ Hence $L\in \odot(ABC)$ $\mathrm{Lemma.1}$ $T,G,H$ are collinear Suppose $AG\cap BC=M$ Note that $AT=BC-2AH=2AM-2AH=2HM$ Hence $\dfrac{AT}{HM}=\dfrac{AG}{GM}=2$ Notice $AT//BC$ So $H,G,T$ are collinear Consider that $\angle BLH=\angle ABC=\angle TCB$ Hence $T,H,L$ are collinear Note that $\angle ABC=180^{\circ}-\angle BDH$ Hence $AB$ is tangent to $\odot(BDH)$ Since $MP^{2}=MB^{2},NQ^{2}=NC^{2}$ Hence $M,N$ lies on the root axis of $\odot(BDH)$,the inscribed circle of $\triangle ABC$ and $\odot(CEH)$,the inscribed circle of $\triangle ABC$ respectively By root axis theorem $MD,NE,HL$ are concurrent Which means that $MD,NE,HG$ are concurrent

KurokoShirai wrote: Let $\Delta ABC$ be an acute-angled triangle with $AB < AC$, $H$ be a point on $BC$ such that $AH\ \bot BC$ and $G$ be the centroid of $\Delta ABC$. Let $P,Q$ be the point of tangency of the inscribed circle of $\Delta ABC$ with $AB,AC$, correspondingly. Define $M,N$ to be the midpoint of $PB,QC$, correspondingly. Let $D,E$ be points on the inscribed circle of $\Delta ABC$ such that $\angle BDH + \angle ABC = 180^{\circ}$, $\angle CEH + \angle ACB = 180^{\circ}$. Prove that lines $MD,NE,HG$ share a common point. Let $W$ be the $A-Why$ point then we have that $(WHDB),(WHEC)$ are concyclic. Also $AB$ is tangent to $(WHDB)$ amd $AC$ is tangent to $(WHEC)$ so $M$ haw equal power on $(WHDB)$ and the incercli similar for $N$ Now by radical center we het that $MD,NE,HG$ share a common point

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