parmenides51 wrote: Let $H$ be a point on the line $AG$ such that $FG \parallel AE$. $H$ is not defined.

Let $I=FH\cap DE$. $Claim: HBIC$ Is cyclic. $Proof:$ Notice that $\angle HID=\angle AED=\angle HGD$, hence $HGID$ is cyclic, so by PoP $FI\cdot FH= FG\cdot FD=FB\cdot FC$ and the claim is proved. Since $DE$ is the perpendicular bisector of $BC$ we have $IB=IC$ and $HF$ angle bisector of $\angle BHC$. Let $H'$ be the orthocenter of $\triangle HAC$, observe $HF\perp AD$, with this we know $A, H, AH'\cap HC, AD\cap HF, HH'\cap AC$ all lie on a circle with diameter $AH$, angle chase to get $\angle BAH'=\angle BAD-\angle H'AD= \angle CAD-\angle CHF=\angle FHH'-\angle FHB=\angle BHH'$ and $AHBH'$ is cyclic.

Consider this configuration, similar reasonings for other cases Let $K$ be the orthocenter of $\triangle HAC, HF\cap \{ AB,AC,DE \}=\{ X,Y,L \}$ Then $AXY$ is isosceles at $A,$ now note that $\angle HGD=\angle AED=\angle HLD\Rightarrow HGLD$ is cyclic $\Rightarrow FH.FL=FB.FD=FB.FC\Rightarrow HBLC$ is cyclic $\Rightarrow \angle BHX=\angle CHY,$ but $\angle BXY=\angle AYH$ $\Rightarrow \angle HBX=\angle ACH\Rightarrow \angle ABH=\angle AKH\Rightarrow K\in (HAB).$