Let us denote the intersection of $AB$ and $DE$ be $K$ , Asume $EL$ be the diameter of $\odot B$ , denote the intersection of $AB$ and $FC$ be $M$ . Join $AL$ , if we can prove $GB // AL$ then since $EB=BL$ ,so $EG=EH$ ,we can finish the proof . Prove : $BG//AL$ : Easy to get :$BC^2=BM*BA=BE^2$ hence :$\triangle BEN \sim \triangle BAE$ because $E,F,M,K$ are concyclic , hence $\angle BEM=\angle FHM=\angle EAB$ hence $FH//EA$ $\Longrightarrow$ $\frac{BF}{EF}=\frac{BH}{HA}$ (#) to $\triangle AFB$ and line $KGE$ , by Menelaus Th. , we get : $\frac{EF}{EB}*\frac{BH}{HA}*\frac{AG}{GF}=1$, By (#) , hence : $\frac{AG}{GF}=\frac{BE}{FB}=\frac{BL}{BF}$ $\Longrightarrow$ $BG//AL$ QED.

i know this may sound stupid, but i used coordinate geometry to solve this problem in 1 hour during the competition

Because I am a little confused with Issl's proof, for example "Let us denote the intersection of $AB$ and $DE$ be $K$" (???) and "$\Delta BEN\sim BAE$" (???) ..., I will post my solution here. However, I see that we both may have some same ideas: Let $FB$ cut the circle with center $B$, radius $BC$ again at $K$, what we have to prove is equivalent to prove that $AK\parallel GB$ Let $BC$ cut the circle with diameter $DB$ again at $I$, as $\angle DIB = 90^{\circ}, DIBE$ is cyclic, hence $\angle EIB = \angle BDE = \angle CFB$, thus $ICEF$ is cyclic, but as $BCE$ is isosceles, it is obvious that $CE\parallel IF$ Thus $\frac{IC}{CB} = \frac{FE}{EB}$, notice that $\frac{IC}{CB} = \frac{DA}{AB}$. Using Melenaus in $\Delta AFB$ we get $\frac{GF}{GA}.\frac{DA}{DB}.\frac{EB}{EF} = 1$, but $\frac{DA}{DB} = \frac{EF}{EB}$, so $\frac{GF}{GA} = \frac{FB}{EB} = \frac{FB}{BK}$ hence $AK\parallel GB$

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Dear lssl: I think ∆$BEM$~∆$BAE$

vslmat wrote: Because I am a little confused with Issl's proof, for example "Let us denote the intersection of $AB$ and $DE$ be $K$" (???) and "$\Delta BEN\sim BAE$" (???) ..., I will post my solution here. However, I see that we both may have some same ideas: Let $FB$ cut the circle with center $B$, radius $BC$ again at $K$, what we have to prove is equivalent to prove that $AK\parallel GB$ Let $BC$ cut the circle with diameter $DB$ again at $I$, as $\angle DIB = 90^{\circ}, DIBE$ is cyclic, hence $\angle EIB = \angle BDE = \angle CFB$, thus $ICEF$ is cyclic, but as $BCE$ is isosceles, it is obvious that $CE\parallel IF$ Thus $\frac{IC}{CB} = \frac{FE}{EB}$, notice that $\frac{IC}{CB} = \frac{DA}{AB}$. Using Melenaus in $\Delta AFB$ we get $\frac{GF}{GA}.\frac{DA}{DB}.\frac{EB}{EF} = 1$, but $\frac{DA}{DB} = \frac{EF}{EB}$, so $\frac{GF}{GA} = \frac{FB}{EB} = \frac{FB}{BK}$ hence $AK\parallel GB$ Obviously you must have misunderstood the statement of this problem. According to the statement, D should be on side AC, not on AB. BTW, I make a solution with projective geometry.

Let $X=DE\cap AB$, $Y=BD\cap CE$, $Z=AB\cap CF$ Notice that $DC$ is also tangent to the circle hence $CE\perp BD$. Now, $\underline{CLAIM.}$$\frac{GX}{XE}=\frac{BX}{BA}$ Proof.Applying Menelaus Theorem to $\triangle BEX$, we have $\frac{XG}{GE}=\frac{AX}{AB}\cdot\frac{FB}{FE}$ It hence suffices to prove $\frac{FB}{FE}=\frac{BX}{AX}$ since$\angle ECF=\angle YCZ=\angle YBZ=\angle DBA$ \begin{align*} \frac{FB}{FE}&=\frac{\sin\angle BCF}{\sin\angle ECF}\cdot\frac{\sin\angle BEC}{\sin\angle CBE}\\ &=\frac{\sin\angle DAB}{\sin\angle ECF}\frac{\sin\angle{BDX}}{\sin\angle ADX}\\ &=\frac{BX}{AX} \end{align*}which is exactly what we want to prove. Hence the claim is proved and $$\frac{GH}{GE}=\frac{GX}{GE}\cdot\frac{GH}{GX}=\frac{BX}{BA}\cdot\frac{BA}{BX}=1$$

Let $P = \overline{AD} \cap \overline{BC}$. Notice that $EDCO$ is cyclic, and furthermore it is the nine-point circle of $ABX$. But $E$ is the $O$-antipode in $(DOCE)$, hence as $F$ is the orthocenter, $M$ is the foot of the $X$-altitude, and thus lies on the circle too.