WHAT A SHAME! It's exactly the same what MEMO 2010, Team Competition, Problem 5 http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2016374&sid=e1fb87e215195f3e82c51d8786915a9f#p2016374

Is the following true? A little angle chasing gives $\angle FHE=\frac{\angle C}{2}$ so $CE=CD=CH$ and so $\angle FBD=\angle DCH....$

Socrates: yes. $\angle DHE=90-\angle EFH=90-\angle GFD-\angle GDE=90-\frac{\angle A}{2}-\frac{\angle B}{2}=\frac{\angle C}{2}$. $\bigtriangleup EDC$ is isoceles $\Rightarrow$ $C$ is the circumcenter of $\bigtriangleup EDH$ $\Rightarrow$ $\bigtriangleup DCH$ and $\bigtriangleup DFB$ are isoceles, so you're done.

Let $DG\cap FE=M$, applying Pascal's to $DDFEEG$ we get $M,C,H$ collinear. Since $FG$ is diameter, applying Pascal's again to $FFDGGE$ where $FF\cap GG=\infty$ we get $MH\parallel FF\Rightarrow CH\parallel AB$.

Dear Mathlinkers, see alos http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=366473 Sincerely Jean-Louis

Alternative solution with Pascal: Let $DG \cap FE = P$, and applying Pascal to $GEEFDD$, we have $P,C,H$ collinear as above. But then we have $DHPE$ cyclic and $HP$ as the diameter, and since $C$ is the intersection of the perpendicular bisector of $ED$ and the diameter of $(DHPE)$, then $C$ is the center of $(DHPE)$, thus $CD = CH$, and we are done by similar triangles.

Better labeling wrote: The incircle of a triangle $ABC$ touches the sides $BC,CA,AB$ at $D,E,F$, respectively. Let $G$ be a point on the incircle such that $DG$ is a diameter. The lines $DE$ and $FG$ intersect at $H$. Prove that $AH\parallel BC$. Redefine $H$, let it be the intersection of $ED$ and line parallel to $BC$ from $A$. We'll show that $F,G,H$ are collinear. We have \begin{align*} \measuredangle AHE=\measuredangle CDE=\measuredangle DEC=\measuredangle HEA\implies AD=AE=AH. \end{align*}Thus, \begin{align*} 2\measuredangle HFA=\measuredangle HAF=\measuredangle CBA=2\measuredangle BDF=2\measuredangle DGF=2\measuredangle GDF=2\measuredangle GFA, \end{align*}we are done. [asy][asy]import olympiad; size(6cm);defaultpen(fontsize(10pt));pen org=magenta;pen med=mediummagenta;pen light=pink;pen deep=deepmagenta;pen dark=darkmagenta;pen heavy=heavymagenta; pair A,B,C,I,D,E,F,G,H; A=dir(120);B=dir(210);C=dir(330);I=incenter(A,B,C);D=foot(I,C,B);E=foot(I,A,C);F=foot(I,A,B); G=2I-D;H=extension(E,D,G,F); draw(A--B--C--cycle,deep);draw(incircle(A,B,C),org); draw(D--G,deep);draw(F--H,deep);draw(D--H,deep);draw(A--H,light); 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("$F$",F,dir(F)); dot("$G$",G,dir(G)); dot("$H$",H,dir(H)); [/asy][/asy]