Let $H \equiv AH_1 \cap BH_2$. Easily $HC \parallel DF \Longrightarrow \measuredangle AFH_2+ \measuredangle DFH_2= \measuredangle AFD = \measuredangle AHC = 90^{\circ} + \measuredangle H_1CH$, $( \star )$. $$\frac{AH}{HH_1} = \frac{AH_2}{H_2D} = \frac{AF}{FD} \cdot \frac{ \sin \angle AFH_2}{ \sin \angle DFH_2} = \frac{AH}{HC} \cdot \frac{ \sin \angle AFH_2}{ \sin \angle DFH_2}$$ $$\frac{ \sin 90}{ \sin \angle H_1CH} = \frac{HC}{HH_1} = \frac{ \sin \angle AFH_2}{ \sin \angle DFH_2} ( \star \star )$$ By $( \star ), ( \star \star )$ and Two Equal Angles Lemma we get $\measuredangle AFH_2 = 90^{\circ}$. Furthermore $FH_2 \parallel BC$.

SOLUTION: Let $H$ be the orthocenter of $\Delta ABC$. Join $HC$ and $H_1$$H_2$. $CH$ being perpendicular to $AB$ is parallel to $DE$. So, $\angle FDH_2$ = $\angle HCH_2$ = $\angle FH_1H_2$. So, $FH_2DH_1$ is cyclic. This implies that $\angle H_1FH_2$ = $90^{\circ}$. Hence, $FH_2 \parallel BC$.

Dear Mathlinkers, we can think to the little Pappus theorem... Sincerely Jean-Lou

Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Orthique%20encyclopedie%200.pdf p. 9... Sincerely Jean-Louis

^yes Pappus works pretty clean over here, Let $H$ be the orthocenter of $\Delta ABC$, then applying Pappus' Hexagon theorem on sets of collinear points $\{ F,H,H_1 \}$ and $\{ C,D,H_2 \}$ $\implies$ $FH_2 \cap H_1C$ lies on the line through $(FD \cap HC)$ and $ (HH_2 \cap H_1D)$ which is the Line at Infinity, hence, $FH_2||BC$