LEMMA(WELL KNOWN) In a $\Delta ABC$ let $\Delta XYZ$ be the anticomplementary triangle and $N$ is the Nagel point of $\Delta ABC$ and $H_1,H_2,H_3$ are the orthocenters of $\Delta BIC,\Delta CIA,\Delta AIB$,then $XH_1,YH_2,ZH_3$ are concurrent at $N$. PROOF Simple homothety. _______________________________________________ Now Let $AI\cap BC=K$ and now note that $ED\parallel AI$,so let a line through $H$ parallel to $AI$ meet $BC,CA,AB$ at $T,D',E'$,now by simple menelaus we have $\frac{BT}{CK}=\frac{BE'}{CA}$ $(\star )$,but by Lemma $H$ lies on $XN$ so $T,K$ are isotomic conjugates wrt $BC$ so $BT=CK$ ,$(\star ),\implies$ $BE'=CA$ $\implies E'=E,D'=D$ so we conclude $H,E,F$ are collinear $\blacksquare$

Just reflect everything across midpoint of $BC$ and realize that you are done.