A conjecture involving three parabolas intrinsically bound to any triangle

Question

Given any triangle $\triangle ABC$, we can build the parabola with directrix passing through the side $AB$ and focus in $C$. This curve intersects the other two sides in the points $D$ and $E$.

Similarly, we can build other two parabolas, one with directrix passing through $AC$ and focus in $B$ (red), and one with directrix passing through $BC$ and focus in $A$ (green), obtaining other $2$ couples of points $F,G$ and $H,I$.

My conjecture is that

The $6$ points $D,E,F,G,H,I$ always determine an ellipse.

How can I show this (likely obvious) result with a simple and compact proof?

Thanks for your help, and sorry for the trivial question!

This problem is related to this one.

Answer 1

We'll take the following Ceva-like result as a given:

For $\triangle ABC$ with $D_B$ and $D_C$ on $\overleftrightarrow{BC}$, $E_C$ and $E_A$ on $\overleftrightarrow{CA}$, and $F_A$ and $F_B$ on $\overleftrightarrow{AB}$, those points lie on a common conic if and only if $$ \frac{BD_B}{D_BC}\cdot\frac{CE_C}{E_CA}\cdot\frac{AF_A}{F_AB} = \frac{CD_C}{D_CB}\cdot\frac{AE_A}{E_AC}\cdot\frac{BF_B}{F_BA} \tag{$\star$}$$

As is typical of Ceva-like results, the individual ratios in $(\star)$ are signed: a ratio is positive if the component directed segments point in the same direction; negative if they point in opposite directions.

Now, to the problem at hand ...

I've renamed points to match the statement above, where a subscript indicates the focus of the parabola through the point. Of course, if a side-line crosses a parabola, then it typically does so again (unless the line is parallel to the axis of the parabola). So, a line-parabola intersection typically consists of two points. For instance, there are two candidate positions where side-line $\overleftrightarrow{CA}$ meets the $C$-focused parabola; I've marked these $E_C^{+}$ and $E_C^{-}$, as the superscripted sign indicates whether the point is on the same side of $C$ as point $A$.

Now, from $E_C^{\pm}$, drop a perpendicular to $C^{\pm}$ on $\overleftrightarrow{AB}$, the directrix of the $C$-focused parabola. By definition of a parabola, $$C^{\pm} E_C^{\pm} = CE_C^\pm \tag{1}$$ Noting that, as an unsigned ratio, $$\frac{C^\pm E_C^\pm}{E_C^\pm A} = \sin A \tag{2}$$ we have the signed ratio $$\frac{CE_C^\pm}{E_C^\pm A} = \pm\sin A \tag{3}$$ Likewise, $$ \frac{AF_A^\pm}{F_A^\pm B}=\pm\sin B \qquad \frac{BD_B^\pm}{D_B^\pm C}=\pm\sin C \tag{4}$$ $$\frac{CD_C^\pm}{D_C^\pm B}=\pm \sin B \qquad \frac{AE_A^\pm}{E_A^\pm C}=\pm\sin C \qquad \frac{BF_B^\pm}{F_B^\pm A}=\pm\sin A$$

Clearly, these give that the ratios on the left- and right-hand sides of $(\star)$ match in absolute value (namely, $\sin A\sin B\sin C$); we make them match completely by choosing appropriate signs. There are $26$ ways to do this, hence $26$ common conics.

In particular, there is a common conic through $D_B^{+}$, $E_C^{+}$, $F_A^{+}$, $D_C^{+}$, $E_A^{+}$, $F_B^{+}$, as per OP's conjecture, and there is a common conic through $D_B^{-}$, $E_C^{-}$, $F_A^{-}$, $D_C^{-}$, $E_A^{-}$, $F_B^{-}$, as per the conjecture I suggested in a comment. $\square$

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Note. This analysis doesn't show specifically that OP's conjectured conic is always specifically an ellipse. ("My" conjectured conic varies in nature. I haven't checked the other $24$.) I'll have to come back to that.