Let $A_1, D_1, E_1 $ be the reflections of $A, D, E $ over $\overline{BC} $. Let $\overline{A_1B} $ intersect $(BDF) $ again at $X $, and $\overline{A_1C} $ intersect $(CEF) $ again at $Y $. $\overline{BF} $ bisects $\measuredangle XBD$ and $BDFX $ - cyclic $\Rightarrow [FD] = [FX].$ $\overline{CF} $ bisects $\measuredangle YCE$ and $CEFY $ - cyclic $\Rightarrow [FE] = [FY].$ $F $ lies on the perpendicular bisector of $[DE] \Rightarrow [FD] = [FE] = [FD_1] = [FE_1] = [FX] = [FY] \Rightarrow D_1E_1XY$ - cyclic. $D_1E_1XY$ - cyclic and $\overline{BC} \parallel \overline{D_1E_1} \Rightarrow BCXY$ - cyclic $\Rightarrow [A_1B] \cdot [A_1X] = [A_1C] \cdot [A_1Y] \Rightarrow A_1 \in \overline{FK}$.

oops I claim $A'$, the reflection of $A$ over $\overline{BC}$, is the fixed point. Vary $D$ linearly as a function of $t \in \mathbb{R}$, and for a point $P$ let $P_x(t)$ denote the $x$-coordinate of its homogenized barycentric coordinate as a function of $t$. Then $E$ varies linearly as well, and by homothety the midpoint of $\overline{DE}$ also varies linearly so $F$ does as well. We now investigate the "degree" of $(BDF)$ in barycentric coordinates. Suppose the circle has equation $-a^2yz-b^2zx-c^2xy+(u(t)x+v(t)y+w(t)z)(x+y+z)=0$; then we have $v(t) \equiv 0$ since $(BDF)$ always passes through $B$. Plugging in $D$ and noting that $D_z(t) \equiv 0$, we have $-c^2D_x(t)D_y(t)+u(t)D_x(t)=0$, so $u(t)=c^2D_y(t)$ varies linearly in $t$. Plugging in $F$ and noting that $F_x(t) \equiv 0$, we have $-a^2F_y(t)F_z(t)+w(t)F_z(t)=0$, so $w(t)=a^2F_y(t)$ varies linearly in $t$ as well, so the circle "varies linearly". The same is true for $(CEF)$. Subtracting the two circle equations from each other and dividing out the $x+y+z$ factor yields the equation of the radical axis of $(BDF)$ and $(CEF)$, i.e. $\overline{FK}$. This line thus varies linearly, since each of its coefficients are formed from subtracting two linear functions from each other. Since $A'$ is fixed, it thus suffices to check two cases. If $D=A$, then $F$ is the foot of the perpendicular from $A$ to $\overline{BC}$, so $K=A$ and the conclusion is clear. If $D=B$, then $F$ is the midpoint of $\overline{BC}$. $(BDF)$ becomes the circle through $F$ tangent to $\overline{AB}$ at $B$, and a similar result holds for $(CEF)$. Then by angle chasing and well-known humpty point properties (or the definition of the humpty point, if you choose to define it as such), it follows that $K$ is the $A'$-humpty point of $\triangle A'BC$, hence the desired collinearity holds as well.

Let $\frac{AD}{AB} = k$. We calculate $P(A) = pow_{(BDF)} (A) - pow_{CEF} (A) = k(c^2 - b^2)$. Let $X$ be the foot from $A$ to $BC$. Then $P(X) = -BX \cdot FX - CX \cdot FX = (-c \cos B - b \cos C) k(XM)$, assuming that $AB < AC$. Now we know that $P$ is linear, so letting $A'$ be the reflection of $A$ over $BC$, we have $P(A') = P(X) + P(X) - P(A) = k (2XM(b \cos C + c \cos B) + (c^2 - b^2))$. We evaluate the inner expression as $(a - 2 c \cos B) (b \cos C + c \cos B) + c^2 - b^2 = (2R \sin A - 4R \sin C \cos B) (2R) (\sin A) + (2R)^2 ((\sin C)^2 - (\sin B)^2) = 0$. Factoring out $4R^2$, we have $\sin^2 A - 2 \sin A \sin C \cos B + \sin^2 C - \sin^2 B = \sin A (\cos C \sin B - \sin C \cos B) + \sin^2 C - \sin^2 B = \sin B (\sin A \cos C - \sin B ) + \sin C (\sin C - \sin A \cos B) = \sin B ( - \sin C \cos A) + \sin C (\sin B \cos A) = 0$, so $P(A') = 0$ and $A'$ lies on the radax of the two circles..

guys, the question is solvable with linearity of pop. what the HE. am i doing (its technically synthetic though! trust!) also why were half the people here competing to solve it first (thonk) Alright this is what we do LMAO. I CANT BELIEVE THIS WORKS I STILL CANT BELIEVE IT. Reflect $A,D,E$ across $BC$ to $A',D',E'$. I claim that $A'$ is the desired point; we just need to show that the power of $A$ to $(BD'F)$ and $(CE'F)$ is equal. Call the circles $\omega_1$ and $\omega_2$. Also drop altitudes $BQ,DR,FP$ on $\triangle BDF$. Let's call $M$ the midpoint of $DE$. \[\text{Pow}_B(\omega_2)-\text{Pow}_B(\omega_1)=BF\cdot BC\]\[\text{Pow}_D(\omega_2)-\text{Pow}_D(\omega_1)=2DF^2-2DQ\cdot DF\]hence by Linearity of PoP we just need to show \[\frac{AD}{AB}\cdot BF\cdot BC=2DF^2-2DQ\cdot DF\]which gives \[2\cdot BF\cdot DM=2DF^2-2DQ\cdot DF\]or \[BF\cdot RF=DF^2-DQ\cdot DF\]which is \[FQ\cdot FD=DF\cdot DF-DQ\cdot DF\]which is true! kudos to #2, honestly need to stop being afraid to draw intersections. i suppose perpendicular means i need to look at internal/external angle bisectors, honestly wish i could have pulled that out of my mind. SUPER clean. congrats!

Let $P$ be the foot from $A$ to $BC$, let $M$ be the midpoint of $BC$, and let $k=\frac{AD}{AB}$. We have $$\text{pow}_{(BDF)}(A) - \text{pow}_{(CEF)}(A) = AD\cdot AB - AE\cdot AC = k\left(AB^2-AC^2\right)$$ $$\text{and}$$ $$\text{pow}_{(BDF)}(P) - \text{pow}_{(CEF)}(P) =-PF\cdot PB - PF\cdot PC = -k\cdot PM(PB + PC) = -k\cdot\frac{1}{2}\left(PC-PB\right)(PB+PC) = \frac{1}{2}k\left(PB^2-PC^2\right)=\frac{1}{2}k\left(AB^2-AC^2\right)$$ so the reflection of $A$ over $P$ always lies on $FK$.

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I hope this works. I desperately wanted to use Steiner conics Work over $\mathbb{RP}^2$. Let $\ell$ be the line through $A$ parallel to $BC$. Let $FK \cap \ell = G$. Finally let $\ell \cap (ABC) = A'$ and let $M$ be midpoint $BC$. Claim : $DG \parallel BA'$. Proof : Since $(BDF)$ and $(CEF)$ intersect at $K$, $K$ also lies on $(ADE)$. Now $\angle GKE=\angle FCE = \angle BCA = \angle GAC = \angle GAE$. Hence $G$ also lies on $(ADEK)$. Hence Homothety at $A$ sending $(ADE)$ to $(ABC)$, sends $D$ to $B$ and $G$ to $A'$, which implies the claim. $\blacksquare$ Vary $F$ on line $BC$. We now show that there is a projective map $f : BC \mapsto \ell$ under which $f(F) = G$. Indeed, $$F \overset{\infty_{\perp BC}}{\underset{BC \mapsto AM}{\mapsto}} F\infty_{\perp BC}\cap DE\cap AM \overset{\infty_{BC}}{\underset{AM \mapsto AB}{\mapsto}}D \overset{\infty_{BA'}}{\underset{AB \mapsto \ell}{\mapsto}}G$$Via the above projective map, $\infty_{BC} \mapsto \infty_{AM} \mapsto \infty_{AB} \mapsto \infty_{\ell} = \infty_{BC}$. Hence $f(\infty_{BC}) = \infty_{BC} $. As $BC \cap \ell$ is $\infty_{BC}$, by steiner conics, $FG$ must pass through a fixed point as desired!

In general the follows holds: Let $ABC$ be a triangle and let $D,E,F$ move linearly with respect to each other on sides $BC,CA,AB$ respectively. Then the radical axis of $(BDF),(CDE)$ passes through a fixed point.

Found a better solution. Since $ BDFK $ and $CEFK $ are cyclic we have that: $\measuredangle DKE = \measuredangle DKF + \measuredangle FKE = \measuredangle DBF + \measuredangle FCE = \measuredangle ABC + \measuredangle BCA = \measuredangle BAC = \measuredangle DAE \Rightarrow ADEK $ is cyclic. Let $\overline{FK} $ intersect the circumcircle of $ADEK $ again at $P $. Since $\overline{DE} \parallel \overline{BC}$ we have that: $\measuredangle ADE = \measuredangle ABC = \measuredangle DKF = \measuredangle DKP.$ This combined with $ADEKP $ being cyclic $\Rightarrow |AE| = |DP| \Rightarrow$ $\Rightarrow APED$ is an isosceless trapezoid. This combined with $|FD| = |FE| \Rightarrow |FA| = |FP|$. This combined with $\overline{AP} \parallel \overline{DE} \parallel \overline{BC} \Rightarrow $ $\Rightarrow \measuredangle BFA = \measuredangle PAF =\measuredangle FPA = \measuredangle PFC \Rightarrow \overline{FA}$ and $\overline{FKP}$ are reflections about $\overline{BC}$. Since $\overline{FA}$ passes through a fixed point its reflection in $\overline{BC}$ also passes through a fixed point.

Taking $D\equiv E \equiv A$ gives the $A$-altitude as the radical axis of the two circles and then an adequate diagram, together with having in mind that the fixed point has to depend only on $A$, $B$, $C$, makes obvious that the desired point is the reflection $A_1$ of $A$ with respect to $BC$. We have to show it has equal powers with respect to the circumcircles of $BDF$ and $CEF$. To compute these powers, let $BA_1$ intersect the circumcircle of $BDF$ and $N$ and $CA_1$ intersect the circumcircle of $BEF$ at $M$. Then the assertion is equivalent to $BCMN$ being cyclic. Note that we can now get rid of $A_1$ and the whole $A$-altitude from the diagram! As a bonus, $\angle NBF = \angle ABC$ and $\angle MCF = \angle ACB$ imply $FN = FD = FE = FM$ from the circumcircles of $BDF$ and $CEF$. All that remains is to angle chase. From $FD = FE$ and $DE \parallel BC$ we have $\angle BFD = \angle EDF = \angle DEF = \angle EFC = t$. With $\angle ABC = \beta$, $\angle ACB = \gamma$ we obtain $\angle BFN = \angle BDN = 180^{\circ} - 2\beta - t$, similarly $\angle CFM = 180^{\circ} - 2\gamma - t$, so $\angle MFN = 2\beta + 2\gamma + 2t - 180^{\circ}$ and $FM = FN$ implies $\angle FNM = 180^{\circ} - \beta - \gamma - t$. Together with $\angle BCM = \gamma $ and $\angle BNF = 180^{\circ} - \angle BDF = \beta + t$ we obtain $\angle BCM + \angle BNM = 180^{\circ}$, which completes the proof.