Let $P \equiv BF \cap CE.$ $\angle (FA,FB)=\angle (DA,DB)=\angle (EA,EC)$ $\Longrightarrow$ $A,E,P,F$ are concyclic. From the complete cyclic $AEPF,$ it follows that the polar of $B$ WRT $\odot(AEF)$ goes through $C$ and vice versa $\Longrightarrow$ $B,C$ are conjugate points WRT $\odot(AEF)$ $\Longrightarrow$ $\odot(AEF)$ is orthogonal to the circle $(M)$ with diameter $\overline{BC}.$ Now all circles through a fixed point $A$ and orthogonal to a fixed circle $(M)$ forms a pencil, the second common point of this pencil is the intersection of $AM$ with the polar of $A$ WRT $(M),$ which in fact is the projection of the orthocenter of $\triangle ABC$ on the median $AM.$

We know that if $BF$ cuts $CE$ at $P$ then $P$ lies on the circle $(AEF)$. Let $H$ be the orthocenter of $\Delta ABC$ with altitudes $AI$ and $BK$, then $B, P, H, C$ are concyclic. Easy to see that the reflection of $A$ over $BC$ lies on this circle $(BHC)$. Notice that both $BDPE$ and $DCFP$ are cyclic. Let $(AEF)$ cuts $(BHC)$ at another point $M$ and the line through $A$ parallel to $BC$ cuts $(AEF)$ at $G$. As $\angle GPF = 180^{\circ} - \angle GAF = \angle BAC = \angle BED = \angle BPD$, $G, P, D$ are collinear. Let $GD, AM$ cut $(BHC)$ at $N, Q$, resp., then $QN\parallel GA\parallel BC$. As $\angle BCN = \angle BPD = \angle ACB$, $N$ is the reflection of $A$ over $BC$ and $N$ lies on the altitude $AI$. Let $AQ$ meets $BC$ at $S$; $I$ is the midpoint of $AN$, $S$ is the midpoint of $AQ$, buts as $BQ = CN = AC$, $S$ is the midpoint of $BC$. As $M$ also lies on $(BHC), M$ is a fixed point on the $A$-median of $\Delta ABC$. (Note: $M$ also lies on the circle $(AHK)$).

Here is a solution that uses $0$ creativity xD $b=AC,c=AB,a=BC$ Let $K$ be on $AC$ such that $\angle KBC=\angle BAC$ and let $X$ be on circle $ABK$ on arc $BK$ without $A$ such that $\frac {BX}{XK}=\frac {b}{c}$. Now by power of point $c\cdot BE=BD\cdot a,CK\cdot b=a^2,CF\cdot b=CD\cdot a$ now $BE=\frac {BD\cdot a}{c}$ and $KF=CK-CF=\frac{BD\cdot a}{b}$ so we get $\frac {BE}{KF}=\frac {b}{c}=\frac {BX}{XK}$ now along with $\angle XBE=\angle XKF$ we have that $BXE\sim XKE$ yeilding $\angle BEX=\angle XFA$ means $AEXF$ is cyclic. The only thing you need to do is notice $K$ when $D=B$ and just check that $\frac {BE}{KF}$ is fixed.

Barycentrics WRT $\triangle ABC$ gives a 5 minute solution. The equation of $(AEF)$ is $-a^2yz-b^2zx-c^2xy+(x+y+z)(ux+vy+wz)=0$ where $u,v,w$ are the powers of $A,B,C$ WRT $(AEF)$, respectively. So $u=0, v=aBD, w=aCD$, by Power of a Point and the cyclic quadrilaterals $AEDB,AFDC$. Now let $y=z=1$ in the equation, which gives $-a^2-b^2x-c^2x+(x+2)a^2=0$ since $a(BD)+a(CD)=a^2$. We get a fixed value for $x$ so we're done.

I will also provide a barycentric coordinate solution. Let $ A = (1, 0, 0), B = (0, 1, 0), C = (0, 0, 1) $ and let $ a = BC, b = CA, c = AB $. Let $ D = (0, r, 1 - r) $ for some real $ r $. We find that the equation of the circumcircle of $ \triangle{ABD} $ is given by: $ a^2yz + b^2xz + c^2xy = (x + y + z)a^2rz $ and it is easy to see that this intersects line $ AC $ at the point $ E = (a^2r : 0 : b^2 - a^2r) $. Similarly we find that $ F = (a^2(1 - r) : c^2 - a^2(1 - r) : 0) $. Now it is easy to find that the circumcircle of $ \triangle{AEF} $ has equation $ a^2yz + b^2xz + c^2xy = (x + y + z)(a^2(1 - r)y + a^2rz) $. It is easy to compute that regardless of the value of $ r $, the point $ (a^2 : b^2 + c^2 - a^2 : b^2 + c^2 - a^2) $ lies on this circumcircle, and moreover lies on the $ A $-median of $ \triangle{ABC} $ so we are done. The motivation for using barycentric coordinates is that despite the large number of circles in this diagram, all are defined by vertices or points on sides of $ \triangle{ABC} $ and so their equations are easy to deal with. Moreover, the condition that our fixed point lies on the relevant median implies that its barycentric coordinates will be nice.

Let $DE,DF\cap (AEF)=E',F'$ again, since $\angle E'DC=\angle A=\angle EF'D$, therefore $EE'\parallel BC$ and $EE'\parallel FF'\parallel BC$ analogously. Let $BE'\cap (AEF)=G$, since $BE*AB=BG*BE'=BD*BC$, $G,E',C,D$ are concyclic $\Rightarrow \angle E'GC=\angle E'DC=\angle 180-\angle F'GE'\Rightarrow G\in CF'$. Now let $AG\cap BC=M$, since $\angle GCM=\angle FF'C=\angle FAG, \angle GBC=\angle BAG$, therefore by similarity $CM^2=GM*AM=BM^2\Rightarrow$ M is the midpoint of $BC$. The length of $GM$ dictates that $G$ is a fixed point and we are done.

My solution: apply an inversion with center $A$ with radius $\sqrt{AB\cdot AC}$ and reflection WRT bisector of $\angle BAC$(assume that $ X'$ is inverse of $X$) then $B\longleftrightarrow C$ under the inversion and $D'$ lies on the arc $ BC$ and $(C, D',E')$ , $(B, D', F')$ are collinear. let $AD'\cap BC=S$ and the tangents from $B,C$ to $\odot (\triangle ABC)$ meet at $R$ because $ E'F'$ is a polar of $S$ ;$E'F'$ passes throw point $R$ which is fixed. So $\odot (\triangle AEF)$ passes throw the inverse of $R$ which lies on the median $M_a$ DONE

Let $\mathcal{I} : X \mapsto X'$ be the composition of an inversion with center $A$ and radius $\sqrt{bc}$, combined with a reflection in the $A$-angle bisector. It is easy to see that $B' \equiv C$ and $C' \equiv B.$ Hence, $\mathcal{I}$ takes line $BC$ to $\omega \equiv \odot (ABC).$ It follows that $D' \in \omega$, and from basic inversive properties (circles through the center of inversion are sent to lines not passing through the center of inversion), we find that $E' \equiv A'B' \cap C'D'$ and $F' \equiv A'C' \cap B'D'.$ Now, note that the image of $\odot (AEF)$ is the line $E'F'$, and the image of the $A$-median is the $A$-symmedian. Therefore, to prove that $\odot (AEF)$ passes through a fixed point on the $A$-median, it suffices to prove that $E'F'$ passes through a fixed point on the $A$-symmedian. To see this, let the tangents to $\omega$ at $X$ (note that $X$ is a fixed point, regardless of where $D'$ lies on $\omega$). We will prove that $X$ is the desired point that we seek. First, recall that $X$ lies on the $A$-symmedian (Lemma 1), as needed. Second, by applying Pascal's Theorem to cyclic "hexagon" $AC'C'D'B'B'$, it follows that $F', X, E'$ are collinear, i.e. $E'F'$ passes through $X.$ This completes the proof. $\square$

Define $\mathbb{P}(P, \Gamma_1, \Gamma_2) = \mathbb{P}(P, \Gamma_1) - \mathbb{P}(P, \Gamma_2)$, where $\mathbb{P}$ denotes the power of a point. It is not hard to check that this function is linear. Let $\omega$ and $\omega_1$ denote the circumcircles of $\triangle ABC$ and $\triangle AEF$. It suffices to show that $\mathbb{P}(M, \omega_1)$ is constant. We compute \begin{align*} \mathbb{P}(M, \omega_1, \omega) &= \frac{1}{2}(\mathbb{P}(B, \omega_1, \omega) + \mathbb{P}(C, \omega_1, \omega)) \\ &= \frac{1}{2}(BE\cdot BA - 0 + CF\cdot CA - 0) \\ &= \frac{1}{2}(BD\cdot BC + CD\cdot BC)\\ &= \frac{BC^2}{2}. \end{align*}So $$\frac{BC^2}{4}-\mathbb{P}(M, \omega_1) = \mathbb{P}(M, \omega_1, \omega)=\frac{BC^2}{2}.$$Therefore, $\mathbb{P}(M, \omega_1)$ is constant as desired.

Here is my solution. I will do bary bash. Actually ELMO 2013 G3 wrote: and that this point lies on the median from $A$ to $BC$. this sentence is just a hint for barycentric. If problem asks to prove that there is fixed point on that circle. The problem would be more difficulter for bary bash. Let $A=(1,0,0)$,$B=(0,1,0)$,$C=(0,0,1)$ Since $D$ is on $BC$ we have $D=(0,t,1-t)$.Intersecting $(ABD)$ and $AC$ we get $F=(a^2t,0,b^2-a^2t)$ Again Intersecting $(ADC)$ and $AB$ we get $E=(a^2(1-t),c^2-a^2(1-t),0)$.Now we have that $(AEF)$ has equation $a^2yz+b^2zx+c^2xy=(x+y+z)(a^2(1-t)y+a^2tz)$ Intersecting median $AM$ of $ABC$ and $(AEF)$ we get $(AEF)\cap AM=(a^2,b^2+c^2-a^2,b^2+c^2-a^2)$ which is fixed. So Done!!!

After a quick glance, it looks like no one else has used this inversion, so I will post my solution.

Dukejukem wrote: we find that $E' \equiv A'B' \cap C'D'$ and $F' \equiv A'C' \cap B'D'.$ Isn't A was sent to a point at infinity? so,what did these intersection mean? thankyou!

Let $M$ be midpoint of $BC$. Denote $BD=t$. We can find that $BF=\frac{at}{c}$, so $AF=\frac{c^2-at}{c}$. Similarly, $AE=\frac{b^2-a(a-t)}{b}$. Let $(AEF)$ intersect A-median at $G$. Let diameter of $(AEF)$ be $1$. By Ptolemy's, $AEsinBAM+AFsinCAM=AGsinBAC$. We know that $\frac{sinBAM}{b}=\frac{sinCAM}{c}=n$. Rearranging, we get $AG=\frac{n(b^2+c^2-a^2)}{sinBAC}$, and hence $G$ doesn't depend on $D$.

Invert around $A$ with radius $\sqrt{bc}$, and reflect the figure around the angle bisector of $\angle A$. Let $X^*$ denote image any point $X$ under this transformation, and let $BB \cap CC=K$ It is required to show there exists a unique point on $A-$symmedian that is always collinear with $E^*$ and $F^*$. By Pascal's theorem on $ABBD^*CC$, $E^*,F^*$ and $K$ are collinear. Since $AK$ is $A-$symmedian, we are done. $\square$

Diagram

Am I missing something because G1 looks harder to me? Anyways... Perform a $\sqrt{bc}$ inversion followed by reflection along the angle bisector of $\angle BAC$. I now claim the desired point is the unique point $X$ on the $A$-symmedian such that $XB, XC$ are tangent to $\omega_{ABC}$ (it's well known that $X$ satisfying this condition lies on $A$-symmedian). Now the inverted problem reads Quote: In $\Delta ABC$, let $D$ be a varible point on the arc $BC$ not containing $A$. $E_D :\overset{\text{def}}{=} BD \cap AC$ and let $F_D :\overset{\text{def}}{=} AB \cap CD$. Let $X$ be the point (on the $A$-symmedian) where the tangents from $B$ and $C$ from $\omega_{ABC}$ meet. Then $X, E_D, F_D$ are colinear Let $G :\overset{\text{def}}{=} AC \cap BD$. By Brocard, $\Delta EFG$ is self-polar w.r.t $\omega_{ABC}$, and hence the pole of $EF$ is $G$. On the other hand, clearly the polar of $X$ w.r.t $\omega_{ABC}$ is $BC$. As clearly $G \in BC$ so we finally get $$G \in BC \Leftrightarrow \text{Pole of EF} \in \text{Polar of X} \overset{\text{La-Hire}}{\Leftrightarrow} X \in EF \Leftrightarrow E, F, X \text{colinear} \blacksquare $$

[asy][asy] unitsize(0.3inches); /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(0cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.746637581395362, xmax = 31.396346604651207, ymin = -14.510337674418569, ymax = 16.20171348837207; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((-2.84,5.09)--(-4.86,-5.25), linewidth(2) + wrwrwr); draw((-4.86,-5.25)--(7.16,-5.21), linewidth(2) + wrwrwr); draw((7.16,-5.21)--(-2.84,5.09), linewidth(2) + wrwrwr); draw(circle((-2.5978372866007646,-0.3246197950741255), 5.420032315860253), linewidth(2) + wrwrwr); draw(circle((3.4073187570611343,1.1509890845253732), 7.385445053925736), linewidth(2) + wrwrwr); draw(circle((-0.3266219383926623,1.8804436497881094), 4.076557523910895), linewidth(2) + wrwrwr); draw(circle((-2.8261034088530366,0.9140743603368577), 4.175948762046959), linewidth(2) + wrwrwr); draw((-5.940877778631267,-10.782810015369954)--(-4.86,-5.25), linewidth(2) + wrwrwr); draw((-2.84,5.09)--(1.15,-5.23), linewidth(2) + wrwrwr); draw((-2.84,5.09)--(-0.30294380615529565,-5.234835087541282), linewidth(2) + wrwrwr); draw((1.3475879470134404,0.7767844145761567)--(-5.940877778631267,-10.782810015369954),linewidth(2) + wrwrwr); draw((-4.410597499008975,-2.9495931384914815)--(4.575253560782421,-2.5477111676058937),linewidth(2) + wrwrwr); /* dots and labels */ dot((-2.84,5.09),dotstyle); label("$A$", (-2.7314839069767416,5.3850788837209285), NE * labelscalefactor); dot((-4.86,-5.25),dotstyle); label("$B$", (-4.739003348837209,-4.952148093023238), NE * labelscalefactor); dot((7.16,-5.21),dotstyle); label("$C$", (7.27615032558141,-4.922185116279052), NE * labelscalefactor); dot((-0.30294380615529565,-5.234835087541282),dotstyle); label("$D$", (-0.2745198139534828,-4.952148093023238), NE * labelscalefactor); dot((-3.863131891834357,-0.14721968394418253),linewidth(4pt) + dotstyle); label("$E$", (-3.7502251162790685,0.08163200000000838), NE * labelscalefactor); dot((2.807261975181314,-0.7266798344367534),linewidth(4pt) + dotstyle); label("$F$", (2.931518697674428,-0.4876645581395254), NE * labelscalefactor); dot((-0.027241070881577307,-2.1851058016296045),linewidth(4pt) + dotstyle); label("$X$", (0.08503590697675023,-1.9558504186046388), NE * labelscalefactor); dot((1.15,-5.23),linewidth(4pt) + dotstyle); label("$M$", (1.2835549767441936,-4.982111069767424), NE * labelscalefactor); dot((1.3475879470134404,0.7767844145761567),linewidth(4pt) + dotstyle); label("$P$", (1.46333283720931,1.0104842790697741), NE * labelscalefactor); dot((-4.410597499008975,-2.9495931384914815),linewidth(4pt) + dotstyle); label("$Q$", (-4.289558697674418,-2.7049248372092887), NE * labelscalefactor); dot((-0.9030289700871873,-2.792721124473026),linewidth(4pt) + dotstyle); label("$D'$", (-0.5741495813953436,-3.5438881860464964), NE * labelscalefactor); dot((4.575253560782421,-2.5477111676058937),linewidth(4pt) + dotstyle); label("$F'$", (4.699334325581407,-2.3154061395348706), NE * labelscalefactor); dot((-5.940877778631267,-10.782810015369954),linewidth(4pt) + dotstyle); label("$E'$", (-6.7165598139534906,-10.375446883720901), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P$ and $Q$ be the feet of the altitudes from $B$ and $C$ to their respective opposite sides in $\triangle ABC$. Let $X=AM\cap (APQ)$ where $M$ is the midpoint of $BC$. It is well known that $(QXMD)$ cyclic, so the inversion $\psi$ at $A$ with power $AQ\cdot AB=AP\cdot AC$ swaps $(X,M)$. Note $\psi(E)=D'P\cap AB\equiv E'$ and similarly $F'\equiv QD'\cap AC=\psi(F)$. Now, we will show that $(AEFX)$ cyclic. Applying $\psi$, we just want to show that $E'MF'$ collinear, but this follows from the fact that $M$ is the pole of $PQ$ wrt $(APQ)$ and Brokard on $APD'Q$.

We will use barycentric coordinates. First, we compute the coordinates of $F$. To do this, we need the equation of $(ABD)$. Let $D=(0,p,q)$ with $p+q=1$. Clearly, since it passes through $A$ and $B$, $u=v=0.$ Plugging in $D$ gives $$wq=a^2pq$$$$w=a^2p.$$Thus, the equation of $(ABD)$ is $$(x+y+z)(a^2pz)=a^2yz+b^2xz+c^2xy.$$Letting $y=0$ (line $AC$), we have $$(x+z)(a^2p)=b^2x.$$Thus, the ratio of $x$ to $x+z$ is $a^2p$ to $b^2$, so we have $$F=(a^2p:0:b^2-a^2p).$$ By symmetry, we have $$E=(a^2q:c^2-a^2q:0).$$ Now, we find the equation of $(AEF)$. Clearly, $u=0$ since it passes through $A$. Plugging in the point $F$, we have $$b^2(w(b^2-a^2p))=b^2a^2p(b^2-a^2p)$$$$w=a^2p.$$Similarly $$v=a^2q.$$Thus, the equation of $(AEF)$ is $$(x+y+z)(a^2qy+a^2pz)=a^2yz+b^2xz+c^2xy.$$We will intersect this with the A-median by setting $y=z=1$. This becomes $$(x+2)(a^2q+a^2p)=a^2+b^2x+c^2x$$$$a^2(x+2)(p+q)=a^2+b^2x+c^2x.$$However, since $p+q=1$, this is just $$a^2x+2a^2=a^2+b^2x+c^2x$$so $$x(a^2-b^2-c^2)=-a^2$$$$x=\frac{a^2}{b^2+c^2-a^2}.$$This means that the intersection of $(AEF)$ with the A-median is $$(\frac{a^2}{b^2+c^2-a^2}:1:1).$$Since this does not depend on $p$ and $q$, this is the desired fixed point, hence done.

We will show the desired point is the $A-Humpty$. Take $\sqrt{bc}$ inversion. Now it is easy to check that the problem turns into the following. Problem. Let $AE'F'$ be a triangle and $B,C$ be points in $AF'$, $AE'$ such that $\angle (BE', CF')=180-\angle BAC$. Show that the intersection of the tangents to $ABC$ at $B$ and $C$ intersects at $E'F'$. This is just straightforward by Brokards Theorem on $ABC$, which ends the problem.

postfarming ig. Lemma: For any $\bullet$, we have \[ \text{Pow}(\bullet, (AEF))-\text{Pow}(\bullet, (AEC))=\text{Pow}(\bullet, (AFB))-\text{Pow}(\bullet, (ABC)). \]Proof. Let $f(\bullet)$ denote LHS and $g(\bullet)$ denote the RHS. Note that $f$ and $g$ are linear functions. It suffices to show that $f(\bullet)=g(\bullet)$ holds for three distinct non-collinear positions of $\bullet$, since every point in the plane can be described a linear combination of those points. This task is easy: choose $\bullet = A, E, F$. It is easy to verify that each of these points all work, by the means of radical axis. It suffices to show that $\text{Pow}(M, (AEF))$ is constant as point $D$ varies, where $M$ is the midpoint of side $BC$. By the lemma, we can reduce this to proving that $\text{Pow}(M, (AEC))+\text{Pow}(M, (AFB))$ is constant. Let $f(\bullet) := \text{Pow}(M, (AEC))-\text{Pow}(M, (ABC))$ and $g(\bullet) := \text{Pow}(M, (AFB))-\text{Pow}(M, (ABC))$. It is well-known that $f$ and $g$ are linear, so \begin{align*} \text{Pow}(M, (AEC))+\text{Pow}(M, (AFB)) &= \frac{f(B)+g(B)+f(C)+g(C)}{2} \\ &= \frac{\text{Pow}(B, (AEC))+\text{Pow}(C, (AFB))}{2} \\ &= \frac{\text{Pow}(B, (ADC))+\text{Pow}(C, (ADB))}{2} \\ &= \frac{BD \cdot BC + DC \cdot BC}{2} \\ &= \frac{BC^2}{2}, \end{align*}which is indeed constant.

We are prompted to invert with respect to $A$ as there are many circles that contain $A$ as a common point. Inverting about $A$, we let $X'$ denote the image of a point $X$ under this inversion. Let $R = AD' \cap B'C'$. Then the polar of $R$ is $E'F'$, and as $B'C'$ passes through $R$, we have the pole of $B'C'$ lies on the polar of $R$ which is $E'F'$. Thus $E', P', F'$ are collinear, where $P'$ is the pole of $B'C'$. Now observe that the pole of $B'C'$ is just $B'B' \cap C'C' = P'$, which is really the symmedian $AP'$ of triangle $AB'C'$. Then under inversion $P' \mapsto P$, as the symmedian is mapped to the median of $A$ under inversion with $A$. Hence $P$ is invariant with regards to $(AEF)$. $\blacksquare$

Very simple with bary + trivial synthetic observations We invoke barycentric coordinates with $\triangle ABC$ as the reference triangle. We let $D = (0, m, n)$ with $m+n = 1$. Notice that since $\triangle ABC \sim \triangle DEB$ we can see that $E = (ma^2:0:b^2-ma^2)$. Similarly $F = (na^2:c^2-na^2:0)$. Now we consider $(AEF)$. We get that $u = 0$, $v = na^2$, and $w = mb^2$. Now we can see that the point $\left(\frac{a^2}{2S_a}:1:1\right)$ always lies on $(AEF)$ and obviously this lies on the median. $\blacksquare$

Perform a force-overlaid inversion about $A$ in $ABC$. Then $ABC$ maps to itself with $B$ and $C$ swapping, and $\overline{BC}$ and $(ABC)$ are swapped. Furthermore, we have that the $M$ of $BC$ maps to $\overline{AM} \cap (ABC)$ under the inversion, so after reflection $M$ maps to the intersection of the $A$-symmedian and $(ABC)$, denoted $M'$ below. Thus, it is equivalent to solve this following: "$ABC$ is a triangle and $D$ is a variable point on $(ABC)$. Let $E=\overline{AB} \cap \overline{CD}$, $F=\overline{AC} \cap \overline{BD}$, $P$ be the intersection of the tangents at $B$ and $C$ to $(ABC)$, and $M'=\overline{AP} \cap (ABC) \neq A$. Prove that $P=\overline{EF} \cap \overline{AM}$ for any choice of $D$." This is sufficient because $P$ reflecting over the $\angle BAC$ bisector maps $P$ to a point on the median, and $P$ is fixed so its image after reflection and inversion must be as well. However, this result follows immediately by Pascal's on $ACCDBB$. $\blacksquare$

Let $M$ be the midpoint of $BC$, $X=AM \cap (AEF)$ such that $X \neq A$. Define $f: \mathbb{R}^2 \rightarrow \mathbb{R}$, such that $f(P)=Pow_{(AEF)}(P)-Pow_{(ABC)}(P)$, where $Pow_\omega (P)$ denotes the power of point $P$ to circle $\omega$. By linearity of pop, we have $2f(M)=f(B)+f(C)$. Computing, we have $f(B)+f(C) = BA \cdot BE + CA \cdot CF = (BD+CD) \cdot BC = BC^2$ and $f(M)=MA \cdot MX + \frac{BC^2}{4}$. It follows that $MX = \frac{BC^2}{4MA}$, so $X$ is a fixed point. $\square$

Define the function $f(X) = \operatorname{pow}(X,A) - \operatorname{pow}(X,(AEF))$ for all points $X$, where $A$ represents the point circle $A$. It suffices to show \begin{align*} f(M) &= \frac 12 f(B) + \frac 12 f(C) \\ &= \frac 12 \left(AB^2+AC^2-BE \cdot BA - CF \cdot CA\right) \\ &= \frac 12 \left(AB^2+AC^2-BD \cdot BC - BE \cdot BC\right) \\ &= \frac 12 \left(AB^2+AC^2-BC^2\right) \end{align*} is fixed, where $M$ is the midpoint of $BC$, as this implies $\operatorname{pow}(M,(AEF))$ is fixed. $\blacksquare$

We claim the desired point is $H_A$, the $A$-Humpty point. Force overlay at $A$ with radius $\sqrt{AB \cdot AC}$. Then $D^*$ is a point on $(ABC)$, $E^* = BD^* \cap AC$, $F^* = CD^* \cap AB$, and $H_A^* = BB \cap CC$. Hence Pascal on $ABBD^*CC$ says $H_A^* \in E^*F^* = (AEF)^*$, as desired. $\blacksquare$

me, a year ago, post 48 wrote: We will use barycentric coordinates. noooooooooo We claim that the answer is the $A$-humpty point, which inverts to the intersection of the tangents to $B$ and $C$. Perform a $\sqrt{bc}$ inversion. The problem now becomes: In triangle $\triangle ABC$, $D$ is a point on the circumcircle. Let $AC$ and $BD$ meet at $E$, and let $AB$ and $CD$ meet at $F$. Let $T$ be the intersection of the tangents at $B$ and $C$. Show that $E,F,T$ are collinear. Let $P$ be the intersection of $AD$ and $BC$. By Brocard, $EF$ is the polar of $P$. Since $P$ clearly lies on the polar of $T$ which is $BC$, $T$ lies on the polar of $P$, which is $EF$, done.

Taking $D$ to be the foot of $A$ altitude, we get that the fixed point, if it exists, must be $A$ humpty point. We now $\sqrt{bc}$ invert. Inverted problem wrote: In cyclic quadrilateral $ABDC$, prove that $CD \cap AB=E, BD \cap AC=F, BB\cap CC=T $ are collinear. Now since the problem is purely projective, the result can follow by numerous ways. I list three: Pascal's on $ABBDCC$. Taking a homography sending $AD \cap BC=X$ to the center of the circle. By brokard's theorem, $EF$ is the polar of $X$ and obviously $BC$ is the polar of $T$ so the required follows by La Hire's Theorem.

We present two approaches. In my opinion it makes sense to just barybash here since it's so clean. We set $A=(1,0,0)$ and etc. Then, let $M$ be the midpoint of $BC$. We know, $M=(0:1:1)$. Let $D=(0:r:s)$ for $r,s\in \mathbb{R}$. We consider the equation of circle $(ABD)$, \[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]Since this circle passes through $A$ and $B$, we have that $u=0$ and $v=0$. Further, this it passes through $D$, \begin{align*} -a^2yz - b^2xz -c^2xy + wz(x+y+z) &= 0\\ -a^2rs &= ws(r+s)\\ w &= \frac{a^2r}{r+s} \end{align*}Now, let $F=(t_1:0:t_2)$. Since $F$ is given to be on $(ABD)$, we also have that \begin{align*} -a^2yz - b^2xz - c^2xy + \frac{a^2r}{r+s}z (x+y+z) &= 0\\ -b^2t_1t_2 + \frac{a^2r}{r+s}t_2(t_1+t_2) &= 0\\ t_2 &= \frac{b^2r-a^2r+b^2s}{a^2r} \end{align*}Thus, $F=(a^2r:0:b^2r-a^2r+b^2s)$. Via an exactly similar calculation, we also have that $E=(a^2s:c^2r+c^2s-a^2s:0)$. Now, we similarly consider the equation of the circle $(AEF)$, \[-a^2yz-b^2xz-c^2xy+(ux+vy+wz)(x+y+z)=0\]Since this circle passes through $A$, $u=0$. Since $F$ must lie on this circle, we have that \begin{align*} -a^2yz-b^2xz-c^2xy + (vy+wz)(x+y+z) &=0\\ -b^2xz + wz(x+z) &= 0\\ b^2x &= w(x+z)\\ a^2b^2r &= w(b^2r+b^2s\\ w &= \frac{a^2r}{r+s} \end{align*}with a similar calculation we can also see that $v= \frac{a^2s}{r+s}$. Now, let $P$ be the intersection of $(AEF)$ and $AM$. Clearly $P=(t:1:1)$. Now, plugging this into our circle equation we have \begin{align*} -a^2-b^2t-c^2t + (v+w)(t+2) &=0\\ -a^2 - b^2t -c^2t + a^2(t+2) &=0\\ t &= \frac{a^2}{b^2+c^2-a^2} \end{align*}Thus, $P=(a^2 : S_A : S_A)$. But clearly, this is a fixed point which does not depend on the position of $D$ along the $BC$ line segment. Thus, the desired result must be true. For those of you allergic to bash, we present the following equally straightforward synthetic solution. Let $P$ denote the intersection of lines $\overline{BF}$ and $\overline{CE}$, and let $H_A$ denote the $A-$Humpty Point in $\triangle ABC$. We can start off by making the following simple observation. Claim : Point $P$ lies on circles $(AEF)$ and $(BH_AC)$. Proof : First of all note that, \[\measuredangle CPB = \measuredangle PCB + \measuredangle CBP = \measuredangle ECD + \measuredangle DBF = \measuredangle EAD + \measuredangle DAF = \measuredangle EAF \]Thus, $\measuredangle EPF = \measuredangle CPB = \measuredangle EAF$, so $P$ lies on $(EAF)$. Further, $\measuredangle CPB = \measuredangle EAF = \measuredangle BAC$ which implies that $P$ lies on $(BHC)$ ($H$ is the orthocenter of $\triangle ABC$). We also know that $H_A$ lies on $(BHC)$ which implies that $P$ lies on $(BH_AC)$ as desired. Now, let $H_A' = (BPC) \cap (AEF)$. Note that, \[\measuredangle BCH_A' = \measuredangle FPH_A' = \measuredangle FAH_A' = \measuredangle CAH_A'\]and similarly, $\measuredangle H_A'BC = \measuredangle H_A'AB$. But, this means $H_A'$ lies on the intersection of the two circles passing through $A$ and tangent to side $BC$ at $B$ and $C$ respectively. We know that this point is simply the $A-$Humpty Point $H_A$ so $H_A' \equiv H_A$. Thus, the point $H_A$ lies on $(AEF)$, implying that the circumcircle of $AEF$ always passes through a fixed point other than $A$, and since it is well known that $H_A$ lies on the $A-$median, we are done.

People would have posted this already multiple times, but I'm gonna do this. Do $\sqrt{bc}$ inversion. Now A, B, C, D lie on the circle Apply pascals on ABBDCC to get that E, F, and the tangents from B and C intersect at a fixed point. The tangents from B and C intersect on the symmedian. When reflected across the angle bisector, this is the median.

Normal inversion suffice: Invert around point $A$ with variable radius. Then: it suffice to show that $E'F'$ passes though a fixed point. As $D'$ varies on $(AB'C'$), let $R = AD'\cap B'C'$. Then $R$ lies on $B'C'$ implies $E'F'$ passes through point $X' = B'B' \cap C'C'$ (due to Brokard's theorem). Thus, $AX'$ is the symmedian of $\triangle AB'C'$ which implies line $AX$ is median of $\triangle ABC$.

Let the circle passing through $A$ and $B$ and tangent to $BC$ meet $(AEF)$ in $X \neq A$. Let $BX$ meet $(AEF)$ in $H \neq X$. Let $EF$ meet $BC$ in $G$. Let $DF $ meet $(AFE)$ in $H'$ Let $EX$ meet $BC$ in $Y$. Let $AX$ meet $BC$ in $M$. Observe that $\angle XBG=\angle XAB=\angle EAX$ $\implies \angle BHA=\angle XHF=\angle XAF= \angle A-x=\angle H'AG-\angle H'BA=\angle BH'A$ $\implies H' \equiv H$ Note that by Power of Point, $BD \cdot BC=BE \cdot BA=BX \cdot BH=$, so $DXHC$ is cyclic. Hence, observe that $\angle A =\angle FAC= \angle HAC=\angle HXC$ Also, $\angle YXC=180-\angle HXC-\angle YXB=180-\angle A-\angle EXH=180-\angle A-(180-\angle XHF-\angle FEH-\angle EHX)=\angle EGC $ (i havent written the full angle chase but its easy and i have outlined it) Hence $EXCG$ is concyclic $\implies BCG=\angle XEF=\angle XAF$ $\implies (AXC)$ is tangent to $BC$ $MB^2=MX \cdot MA=MC^2$ $\implies M$ is the midpoint of $BC$, and the proof follows upon noting that $X$ is fixed for a fixed triangle. $\blacksquare$