This problem is just a combination of well known stuff. Note: Throughout the solution all lengths and angles are oriented. Lemma: If $A,C,B,D$ are points on a line and $(A,B;C,D)=-1$, then $DC\cdot DM=DA\cdot DB$, where $M$ is the midpoint of $\overline{AB}.$ Proof: $DM\cdot DC=DM^2-MD\cdot MC=DM^2-\left(\frac{AB}{2}\right)^2=DA\cdot DB.$ Back to the original problem, let $B=ED\cap GF, X=CH\cap ED,Y=CH\cap GF$ and let $J_1,J_2$ be the midpoints of $\overline{ED}$ and $\overline{CH}.$ By the lemma we know that $BX\cdot BJ_1=BE\cdot BD=BG\cdot BF=BY\cdot BJ$, so $XJ_1JY$ is concyclic. Taking into consideration that $J,J_1,J_2$ are collinear (Newton line) we obtain $\angle{J_2XI}=\angle{JXI}=\angle{BXY}=\angle{J_1JB}=\angle{J_2JI}$, so $J_2XJI$ is cyclic. Therefore $YJ\cdot YI=YX\cdot YJ_2=YH\cdot YC$ (the last equality is by the lemma), which implies the conclusion.

@above : Nice observation !

Similar problem https://artofproblemsolving.com/community/c6h355791p1932936

This might be a stupid question but I tried to find which quadrilateral corresponded to the Newton Line $J_1-J_2-J$ in hurricane's solution, and was unable to find it. Maybe I have a configuration issue??? Or maybe I'm just not seeing the quadrilateral. If someone could help me clarify which quadrilateral he's talking about, that would be awesome.

Plops wrote: This might be a stupid question but I tried to find which quadrilateral corresponded to the Newton Line $J_1-J_2-J$ in hurricane's solution, and was unable to find it. Maybe I have a configuration issue??? Or maybe I'm just not seeing the quadrilateral. If someone could help me clarify which quadrilateral he's talking about, that would be awesome. The quadrilateral formed by lines $DF$,$DG$,$EF$,$EG$