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When playing around with conics in GeoGebra, I have found out that the following relation seems to hold:

Let parabola $p$ be tangent to sides/extensions of sides $BC,CA,AB$ of triangle $ABC$ at points $P,Q,R$. Call $F$ the focus of this parabola. Let $T$ be intersection point of lines $AP,BQ,CR$ (they are concurrent as a corollary to Brianchon theorem) and let $S$ be the intersection point of circumcircle and Steiner's circumellipse other than $A,B,C$ (a.k.a. Steiner point). Then $F,T,S$ are collinear.

My question is: can anyone provide a proof of this relation? I imagine this would be quite a complex result to show, but maybe it already exists somewhere in literature (in which case I'd be thankful for a reference).

Thanks in advance.

Illustration: enter image description here

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I do not know a reference nor an enlightening synthetic reason, but a coordinate-based proof is quite easy.

Here I use barycentric coordinates with reference triangle $ABC$. Let its side lengths be denoted $a,b,c$. Points are represented as $(X:Y:Z)$, straight lines as $[U:V:W]$, with a point being on a line if and only if $UX+VY+WZ=0$. In the barycentric system, $[1:1:1]$ represents the line at infinity.

I will not show the derivations of the equations below, but I intend to give enough detail for verification.

First, I will suppose that $ABC$ is not equilateral, in order to have a uniquely defined Steiner point $S$.

Equation of the circumcircle: $$a^2 YZ + b^2 ZX + c^2 XY = 0$$ Equation of the Steiner circumellipse: $$YZ + ZX + XY = 0$$ Steiner Point (located on both the above): $$S = \Bigl((c^2-a^2)(a^2-b^2):(a^2-b^2)(b^2-c^2):(b^2-c^2)(c^2-a^2)\Bigr)$$ Let $E=(X_E:Y_E:Z_E)$ be the point at infinity indicating the direction of the symmetry axis of the parabola $p$. As $E$ lies on the line at infinity, it must satisfy $$X_E + Y_E + Z_E = 0$$ Straight lines $[U:V:W]$ are tangent to the parabola $p$ if and only if $$X_E VW + Y_E WU + Z_E UV = 0$$ In other words, the dual parabola can be represented with coefficient matrix $$p^* = \begin{bmatrix} 0 & Z_E & Y_E \\ Z_E & 0 & X_E \\ Y_E & X_E & 0 \end{bmatrix}$$ Its columns also give the coordinate triples of the points $P,Q,R$.

As a consequence of the focal property, the focus $F=(X_F:Y_F:Z_F)$ is the isogonal conjugate of $E$: $$E = \left(\frac{a^2}{X_F}:\frac{b^2}{Y_F}:\frac{c^2}{Z_F}\right)$$ The condition that $E$ be a point at infinity thus turns into $$\frac{a^2}{X_F} + \frac{b^2}{Y_F} + \frac{c^2}{Z_F} = 0$$ which says that $F$ lies on the circumcircle. This is known as Lambert's theorem.

The Brianchon point $T=(X_T:Y_T:Z_T)$ turns out to be $$T = \left(\frac{1}{X_E}:\frac{1}{Y_E}:\frac{1}{Z_E}\right) = \left(\frac{X_F}{a^2}:\frac{Y_F}{b^2}:\frac{Z_F}{c^2}\right)$$ which makes it the isotomic conjugate of $E$. That relationship can easily be demonstrated in a coordinate-free way, but it is most obvious in barycentric coordinates. With those, we can also verify that $T$ lies on the Steiner ellipse.

Now we are ready to check whether $F,T,S$ are collinear: $$\begin{align} \langle F,T,S\rangle &= \begin{vmatrix} X_F & X_T & X_S \\ Y_F & Y_T & Y_S \\ Z_F & Z_T & Z_S \end{vmatrix} = \begin{vmatrix} X_F & \frac{X_F}{a^2} & (c^2-a^2)(a^2-b^2) \\ Y_F & \frac{Y_F}{b^2} & (a^2-b^2)(b^2-c^2) \\ Z_F & \frac{Z_F}{c^2} & (b^2-c^2)(c^2-a^2) \end{vmatrix} \\ &= \frac{X_F Y_F Z_F}{a^2 b^2 c^2} \begin{vmatrix} 1 & b^2 c^2 & (c^2-a^2)(a^2-b^2) X_F^{-1} \\ 1 & c^2 a^2 & (a^2-b^2)(b^2-c^2) Y_F^{-1} \\ 1 & a^2 b^2 & (b^2-c^2)(c^2-a^2) Z_F^{-1} \end{vmatrix} \\ &= \frac{X_F Y_F Z_F}{a^2 b^2 c^2}(b^2-c^2)(c^2-a^2)(a^2-b^2) \left(\frac{a^2}{X_F} + \frac{b^2}{Y_F} + \frac{c^2}{Z_F}\right) = 0 \end{align}$$ So, yes, those points are collinear.

Disregarding the parabola-related stuff, this finding basically says that, when given a point at infinity, then its isogonal conjugate and its isotomic conjugate are collinear with the Steiner point.

If $ABC$ is equilateral, then there is no unique Steiner point, but the above formulae show that then $F$ coincedes with $T$.

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