Let $Q,R$ be the projections of $B,C$ on $AC,AB.$ $H \equiv BQ \cap CR$ is the orthocenter of $\triangle ABC.$ Inversion with center $A$ and power $AB \cdot AR=AC \cdot AQ$ takes $BC$ into $\omega_A \equiv \odot(AQHR)$ and $\odot(ABE),$ $\odot(ACD)$ into the the tangents of $\omega_A$ at $R,Q.$ They pass through the midpoint $M$ of $BC$ (well-known) and cut $AC,AB$ at the inverses $E',D'$ of $E,D.$ Thus $DE$ is taken into $\odot(AD'E')$ cutting $\omega_A$ again at the inverse $F'$ of $F.$ Now from the problem Concyclic Quadrilateral, $F'$ is on Euler line $OH$ of $\triangle ABC,$ i.e. $OHF' \perp AF'F,$ as desired.

Let $ O, $ $ H $ be the circumcenter, orthocenter of $ \triangle ABC $, respectively. Let $ \odot (H,HA) $ cuts $ CA, $ $ AB $ again at $ Y, $ $ Z $, respectively. Clearly, $ YZ $ is antiparallel to $ BC $ WRT $ \angle BAC $, so $ B, $ $ C, $ $ Y, $ $ Z $ lie on a circle $ \Omega $. From $ \angle DCB $ $ = $ $ \angle BAC $ $ = $ $ \angle CZB $ we get $ CD $ is tangent to $ \Omega $ at $ C $. Similarly, we can prove $ BE $ is tangent to $ \Omega $ at $ B $, so from Pascal theorem (for $ BBCCYZ $) $ \Longrightarrow $ $ F $ lies on $ YZ $, hence $ F $ is the radical center of $ \Omega, $ $ \odot (O), $ $ \odot (H,HA) $ $ \Longrightarrow $ $ AF $ is the radical axis of $ \odot (O) $ and $ \odot (H,HA) $ $ \Longrightarrow $ $ AF $ is perpendicular to the Euler line $ OH $ of $ \triangle ABC $.

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Let $G, H, N$ be the centroid, orthocenter, nine-point center of $\triangle ABC$, respectively. Let $\triangle M_aM_bM_c$ be the medial triangle and $\triangle H_aH_bH_c$ be the orthic triangle. Let $Y, Z$ be the reflections of $A$ in $H_b, H_c$ respectively. Denote $P \equiv M_bM_c \cap H_bH_c.$ Lemma: $AP$ is perpendicular to the Euler line $e.$ Proof: Take $Q \equiv M_bH_c \cap M_cH_b.$ Then $Q \in e \equiv GH$ by Pappus' Theorem for $M_bBH_bM_cCH_c.$ Hence, $e \equiv QN$, which is perpendicular to $AP$ by Brokard's Theorem for $M_bM_cH_bH_c.$ $\blacksquare$ Dilating the above result with center $A$ and ratio $2$, it follows that the line connecting $A$ and $F^* \equiv YZ \cap BC$ is perpendicular to $e.$ We will show that $F^* \equiv F.$ Note that $\triangle ABY \sim \triangle ZCA.$ Moreover, $\measuredangle BCA = \measuredangle CDB = \measuredangle CDZ$, which entails $ABYC \sim ZCAD.$ Therefore, $CA / CY = DZ / DA.$ Similarly, $BA / BZ = EY / EA.$ Thus, Menelaus' Theorem for $\triangle AYZ$ cut by $\overline{DEF}$ and $\overline{BCF^*}$ yields $FY / FZ = F^*Y/ F^*Z$, i.e. $F^* \equiv F$, as desired. $\square$

Here is my solution with complex bashing. Let $x,y,z$ denote the length of $\overline{AB},\overline{BC},\overline{CA}$, and $a,b,c$ be the coordinate of $A,B,C$ respectively. Applying Menelaus's Theorem from $DEF$ and $\triangle ABC$ yields that \[\frac{BD}{DA}\cdot\frac{AE}{EC} \cdot\frac{CF}{FB}=1. \qquad (1)\] Since $\overline{BC}$ is tangent to $\odot(ADC)$, so $BC^2=BD\cdot BA \implies BD=\dfrac{y^2}{x}, DA=\dfrac{x^2-y^2}{x} \implies \dfrac{BD}{DA}=\dfrac{y^2}{x^2-y^2}$, similarly $\dfrac{AE}{EC}=\dfrac{y^2-z^2}{y^2}$. Combining $(1)$ we get $\dfrac{CF}{FB}=\dfrac{x^2-y^2}{y^2-z^2}$. Clearly $x^2=(a-b)(\overline{a}-\overline{b})=2-\dfrac{a^2+b^2}{ab}$, similarly $y^2=2-\dfrac{b^2+c^2}{bc}, z^2=2-\dfrac{c^2+a^2}{ac}$. Thus we can calculate the coordinate of $F$ \[f=\frac{x^2-y^2}{x^2-z^2}b+\frac{y^2-z^2}{x^2-z^2}c=\frac{(a-c)(b^2-ac)}{(b-c)(a^2-bc)}b+\left(1-\frac{(a-c)(b^2-ac)}{(b-c)(a^2-bc)}\right)c.\qquad (2)\] Define $f'=kb+(1-k)c$ is a point on $\overline{BC}$ such that $AF'\perp OH$, where $k$ is a constant. It suffices to prove that $f\equiv f'$. We have \begin{align*} AF'\perp OH \iff 0 &=\frac{a-f'}{a+b+c}+\overline{\left(\frac{a-f'}{a+b+c}\right)}\\ &=\frac{a-kb-c+kc}{a+b+c}+\frac{\frac{1}{a}-\frac{k}{b}-\frac{1}{c}+\frac{k}{c}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\\ &=\frac{(a-kb-c+kc)\cdot \left(ab+bc+ca\right)}{(ab+bc+ca)(a+b+c)}+\frac{(bc-kac-ab+kab)\cdot(a+b+c)}{(ab+bc+ca)(a+b+c)}\\ &=\frac{k(a^2b-a^2c+bc^2-b^2c)-(ab^2+ac^2-b^2c-a^2c)}{(a+b+c)(ab+bc+ca)}\\ &=\frac{k(a^2-bc)(b-c)-(b^2-ac)(a-c)}{(a+b+c)(ab+bc+ca)} \end{align*}Hence $k=\dfrac{(b^2-ac)(a-c)}{(a^2-bc)(b-c)}$. Compare with $(2)$ we conclude that $f\equiv f'$, as desired. We're done. A diagram below: [asy][asy] import graph; import olympiad; import cse5; defaultpen(fontsize(10pt)); size(8cm); pair A,B,C,D,E,F,O,H; path cir1; A=dir(55); B=dir(-150); C=dir(-30); O=origin; H=IP(foot(A,B,C)--A,B--foot(B,A,C)); F=IP(B--C,L(A,foot(A,O,H))); cir1=CP(extension(C,(B-C)*dir(-90)+C,O,foot(O,A,C)),A); D=OP(cir1,B--A); E=extension(F,D,C,A); draw(A--B--C--cycle); draw(A--F); draw(O--H); draw(E--F,blue+dashed); draw(unitcircle); draw(cir1,red+linetype("4 4")); draw(circumcircle(A,B,E),red+linetype("4 4")); draw(C--D); draw(A--E); dot("$E$",E,dir(E)); dot("$A$",A,dir(A)*1.5); dot("$B$",B,dir(B)*1.5); dot("$C$",C,dir(C)*1.5); dot("$D$",D,dir(120)); dot("$F$",F,dir(F)); dot("$O$",O,dir(-90)); dot("$H$",H,dir(-90)); [/asy][/asy]

Notice that the problem is equivalent to showing that line $AF$ is the radical axis of the circles with diameters $\overline{AH}$ and $\overline{AO}$. Consider the transformation which is a composition of inversion about $A$ with power $r^2=AB\cdot AC$, reflection about the $A$-angle bisector, and dilation from $A$ with ratio $\tfrac{1}{2}$. We can now recast the problem as the following: In triangle $ABC$, suppose the tangents to $\odot(ABC)$ at $B$ and $C$ intersect $AC$ and $AB$ at $D$ and $E$, respectively. Furthermore, the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect $AC$ and $AB$ at $P$ and $Q$, respectively. If $\odot(ADE)$ intersects $\odot(ABC)$ again at $F$, show that lines $AF, PQ$, and $BC$ are concurrent. Since $\angle BPC=180^{\circ}-2\angle BAC=\angle BQC$, it follows that $\odot(BQPC)$ is cyclic, and $X\equiv BD\cap CE$ lies on this circle. By angle chasing, we get that $PX\parallel BC$ and $QX\parallel AC$. Then $$(D,C;P,P_{AC\infty})\stackrel{X}{=}(B, E; P_{AB\infty},Q)\Longrightarrow \frac{PC}{PD}=\frac{QB}{QE} $$which implies that $F$ is the spiral center sending $DE$ to $PQ$, hence $\odot(AFPQ)$ is cyclic. The conclusion now follows from the radical axis theorem on $\odot(ABC), \odot(BPQC)$, and $\odot(AFPQ)$.

Lemma: Let $M$ , $N$ be the midpoints of sides $AB$ , $AC$ in a given triangle $ABC$ and $BE$ and $CF$ be its altitudes. then if $EF\cap MN=T$ prove that $AT$ is perpendicular to the Euler line of $ABC$. Proof: Let $R$ be the nine point center of $\triangle ABC$ then $EFMN$ is cyclic quadrilateral with center $R$. Let $S\in AT$ be the miquel point of $EFMN$. so: $$RS\perp AT\ ,\ \angle HSA=90^\circ\Longrightarrow S\in OH\Longrightarrow OH\perp AT$$Back to the problem , Let $N,P$ be the midpoints of $AC,AB$ respectively and $BH_B,CH_C$ be the altitudes of $ABC$. $DC\cap BE=X,NH_C\cap PH_B=Y,H_BH_C\cap NP=S$. From the Lemma $AS\perp OH$. so it's enough to prove that $A,S,F$ are collinear ; since $AS,AF$ are harmonic conjugates of $AY,AX$ WRT lines $(AB,AC)$ it's enough to prove that $A,X,Y$ are collinear. Note that $XB,XC$ are tangents of $\odot(BHC)$ , so if the tangents from $M,N$ to the nine point circle of $ABC$ intersect at $X'$ , $X$ is image of $X'$ under the dilation $H(A,2)$. on the other hand from pascal's theorem for $NNH_CPPH_B\Longrightarrow A,X',Y$ are collinear. Hence $A,X,Y$ are collinear. 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draw((11.81606600170957,7.126589758293002)--(19.977508095641813,10.484251929653063), linewidth(0.8)); draw((3.856742200217277,9.446767090999579)--(10.033298393671755,9.844271989176846), linewidth(0.8)); draw((10.033298393671755,9.844271989176846)--(19.977508095641813,10.484251929653063), linewidth(0.8)); draw((4.898637863432866,17.671638619429658)--(8.652093840198047,6.891688770219023), linewidth(0.8) + linetype("4 4")); draw((3.856742200217277,9.446767090999579)--(6.775365851815459,12.281663694824351), linewidth(0.8)); draw((6.775365851815459,12.281663694824351)--(10.033298393671755,9.844271989176846), linewidth(0.8)); draw((4.898637863432866,17.671638619429658)--(35.056378327850716,3.296865239876465), linewidth(0.8) + linetype("4 4")); draw(shift((9.457085029421917,-5.61651764000664))*xscale(9.533269497887732)*yscale(9.533269497887732)*arc((0,0),1,53.19834500811046,134.1662725316145), linewidth(0.8)); draw(shift((2.2726034159026005,9.647439809546482))*xscale(8.442974799213022)*yscale(8.442974799213022)*arc((0,0),1,-86.31769123013748,71.8785717532574), linewidth(0.8)); 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label("$S$", (20.077182569102394,10.606889736119191), NE * labelscalefactor); dot((8.203649118309867,8.179624912809405),linewidth(2.pt) + dotstyle); label("$Y$", (7.3348277912996105,8.280024950607386), NE * labelscalefactor); dot((8.652093840198047,6.891688770219023),linewidth(2.pt) + dotstyle); label("$X$", (7.778040131397098,6.645679446497901), NE * labelscalefactor); dot((6.775365851815459,12.281663694824351),linewidth(2.pt) + dotstyle); label("$X'$", (5.8389861434705885,12.324337553996953), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy]

This problem in the link can be considered as an extension of this original problem https://artofproblemsolving.com/community/q1h1198358p5970166 Let $ABC$ be a triangle and $D$ is a fixed point on $BC$. $P$ is a point such that $AP\perp BC$. $K$ is projection of $P$ on $AD$. Circumcircles of triangles $KAB,KAC$ cut $CA,AB$ again at $E,F$, reps. $EF$ cuts $BC$ at $Q$. Prove that the line passes through $P$ and is perpendicular to $AQ$ always passes through a fixed point when $P$ moves. Proof by Telv Cohl in the same link.

Dear Mathlinkers for the initial problem http://jl.ayme.pagesperso-orange.fr/Docs/3.%20Perpendiculaires%20a%20la%20droite%20de%20Euler.pdf p. 11... Sincerely Jean-Louis