Let $X = (-1:1:1)$ be the point such that $ABXC$ is a parallelogram. Claim: Points $D$, $E$, $X$ are collinear. Proof. In barycentrics with reference $ABC$, we have $D = (0:s-c:s-b)$, $E = (a-b-c:2b:2c)$ and $X=(-1:1:1)$. Then \[ \det \begin{bmatrix} 0 & s-c & s-b \\ a-b-c & 2b & 2c \\ -1 & 1 & 1 \end{bmatrix} = (a-b-c)[(s-b)-(s-c)]-[2c(s-c)-2b(s-b)] = 0 \]as needed. $\blacksquare$ [asy][asy] pair A = dir(110); pair B = dir(190); pair C = dir(350); pair I = incenter(A, B, C); pair D = foot(I, B, C); pair X = B+C-A; pair H = orthocenter(A, B, C); pair F = foot(H, D, X); pair E = extension(A, I, D, X); filldraw(unitcircle, invisible, blue); draw(A--B--C--cycle, blue); draw(B--X--C, blue); filldraw(circumcircle(B, X, C), invisible, deepgreen); draw(H--F--X, red); draw(A--E, deepgreen); draw(I--D, lightblue); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$I$", I, dir(160)); dot("$D$", D, dir(250)); dot("$X$", X, dir(X)); dot("$H$", H, dir(130)); dot("$F$", F, dir(F)); dot("$E$", E, dir(230)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ A = dir 110 B = dir 190 C = dir 350 I 160 = incenter A B C D 250 = foot I B C X = B+C-A H 130 = orthocenter A B C F = foot H D X E 230 = extension A I D X unitcircle / 0.1 lightcyan / blue A--B--C--cycle / blue B--X--C / blue circumcircle B X C / 0.1 lightgreen / deepgreen H--F--X / red A--E / deepgreen I--D / lightblue */ [/asy][/asy] On the other hand, the points $B$, $H$, $X$, $C$ are well known to be cyclic on the circle with diameter $\overline{HX}$. As $\angle HFX = 90^{\circ}$, point $F$ lies on this circle as well,

Let $G$ be the antipode of $D$ wrt the incircle of $\triangle ABC$, $M$ be the midpoint of $\overline{BC}$, and $D'$, $A'$ be the reflections of $D$ and $A$ over $M$, respectively (note that $D'$ is the $A$-extouch point of $\triangle ABC$). Claim: $D$, $E$, and $A'$ are collinear Proof: By a homothety at $A$, we find that $A$, $G$ and $D'$ are collinear. Because $M$ is the midpoint of $\overline{DD'}$ and $\overline{AA'}$, we find $ADA'D'$ is a parallelogram, so $\overline{AGD'} \parallel \overline{DA'}$. However, because $I$ is the midpoint of $\overline{AE}$ and $\overline{DD'}$, we also find that $ADED'$ is a parallelogram and $\overline{AGD'} \parallel \overline{DE}$, so indeed $D$, $E$ and $A'$ are collinear. Now, note that by reflecting over $M$, we know that $A'$ is the antipode of $H$ wrt $(BHC)$. Since $\overline{HF} \perp \overline{FDEA'}$, we find that $F$ lies on $(BHC)$ and we are done.

The problem is equivalent to $DE$ passing through the antipode of $H$ on $(BHC)$, $H'$ However, this antipode is just the reflection of $A$ across the midpoint of $BC$. Using barycentric coordinates, $D=(0:s-c:s-b)$ and $E=2I-A=(a-b-c:2b:2c)$ and $H'=(-1,1,1)$. However, $$ \begin{vmatrix} 0 & s-c & s-b \\ a-b-c & 2b & 2c\\ -1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} 0 & s-c & s-b \\ 0 & a+b-c & a-b+c\\ -1 & 1 & 1 \end{vmatrix} $$ but the second row is now twice the first row, so this evaluates to 0 and we are done.

FIRE Reflect over $BC$, sending $h$ to $H'$ and $F$ to $F'$. I claim that $F'$ is the mixtilinear intouch point, which finishes as $(BHC)$ and $(ABC)$ are reflections. Let $T$ be the mixtilinear point. Claim: Line $TD$ passes through $A'$ with $AA'\parallel BC$. Proof (by OronSH) Let $J$ be the extouch point, then $\angle BA'D=\angle BA'T=\angle BAT=\angle CAJ$ and symmetry finishes. $\blacksquare$ Thus $TD$ passes through the antipode of $H'$, so it suffices to show that lines $DT$ and $DE$ are reflections about $BC$ Claim: Circle $(DEI)$ is tangent to line $TI$. Proof. Let $D'$ be the top on the incircle and $N$ be the midpoint of arc $BAC$, and $M$ be the other arc midpoint. Reflect over $I$, so this becomes $ADI'$ is tangent to $TI$. This is true as \[\angle IAD'=\angle IAT=\angle MAT=\angle MNT=\angle MNI=\angle NID'\]since $MN \parallel D'I$ $\blacksquare$. Claim: Circle $(TDI)$ is tangent to line $AI$. Proof. Note that by Reim $ND'$, $(ABC)$, and $(AD'I)$ concur at $K$. Thus \[\angle DIE=\angle D'IA=\angle D'KA=\angle NKA=\angle A'TN=\angle DTI\]where $\angle NKA=\angle A'TN$ since $N$ is the top of the circle and $AA'\parallel BC$ $\blacksquare$ Remark. In fact, the last two claims give that $D$ is the $I$ dumpty point of triangle $TIE$.

Very nice problem. Let $M$ be the midpoint of $\overline{BC}$. Reflect $A$ through $M$ to get $A'$. We see that $(A'HBC)$ is cyclic, since \[ \angle BA'C + \angle BHC = \angle BAC + 180 - \angle BAC = 180. \]Moreover $A'H$ is the diameter of this circle, since $ BH \perp AC \parallel A'B$. Hence it is sufficient to prove that $\overline{A'FED}$ are collinear. Define $D'$ as the relfection of $D$ through $M$ and $P$ as the reflection of $D$ through $I$. It is known that $\overline{APD'}$ are collinear. Reflecting $A$ and $D'$ through $M$ we get that $A'D \parallel APD'$. Next reflecting $A$ and $P$ through $I$ we get that $DE \parallel APD'$. The desired result clearly follows. $\square$