Csdrhrt

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.

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
$$ frac{BD_B}{D_BC}cdotfrac{CE_C}{E_CA}cdotfrac{AF_A}{F_AB} =
frac{CD_C}{D_CB}cdotfrac{AE_A}{E_AC}cdotfrac{BF_B}{F_BA} tag{$star$}$$

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
$$frac{CE_C^pm}{E_C^pm A} = pmsin A tag{3}$$
Likewise,
$$
frac{AF_A^pm}{F_A^pm B}=pmsin B qquad
frac{BD_B^pm}{D_B^pm C}=pmsin C tag{4}$$
$$frac{CD_C^pm}{D_C^pm B}=pm sin B qquad frac{AE_A^pm}{E_A^pm C}=pmsin C qquad frac{BF_B^pm}{F_B^pm A}=pmsin A$$

Clearly, these give that the ratios on the left- and right-hand sides of $(star)$ match in absolute value (namely, $sin Asin Bsin C$); 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. $square$

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.

Here is a proof using barycentric coordinates. For non-degenerate triangle $ABC$, we are going to show the six points $D, E, F, G, H, I$ lies on an ellipse.

Let

• 
  $a = |BC|, b = |CA|, c = |AB|$. WOLOG, we will assume $a ge b, c$.


• 
  $R$ and $Delta$ be the circumradius and area of $triangle ABC$.


• 
  $alpha = frac{sin A}{1 + sin A} = frac{a}{2R+a}$,
  $beta = frac{sin B}{1 + sin B} = frac{b}{2R+b}$ and
  $gamma = frac{sin C}{1 + sin C} = frac{c}{2R+c}$
  

For any point $P$ and line $ell$, let $(u_P, v_P, w_P)$ be its barycentric coordinates with respect to $triangle ABC$ and $d_ell(P)$ be the distance between $P$ and line $ell$.

Consider point $D$. Since it lies on $AC$, $(u_D, v_D, w_D)$ has the form $(u, 0, 1-u)$. Since $D$ lies on the parabola having $C$ as focus and $AB$ as directrix, we have

$$bu = |CD| = d_{AB}(D) = d_{AB}(C)(1-u) = frac{2Delta}{c}(1-u) = bsin A(1-u)$$

This implies $u = frac{sin A}{1 + sin A}$ and hence $D$ lies on the line $u = alpha$ in barycentric coordinates.

By a similar argument, we have
$G$ lies on line $u = alpha$, $E$, $H$ lie on line $v = beta$ and
$F$, $I$ lie on line $w = gamma$.
As a result, the six points $D, E, F, G, H, I$ lies on the cubic

$$(u - alpha)(v - beta)(w - gamma) = 0$$
in barycentric coordiantes. These six points belong to another cubic $uvw = 0$ which corresponds to the three lines supporting the triangle. So they belong to the "difference"

$$mathcal{Q}(u,v,w) stackrel{def}{=} uvw - (u-alpha)(v-beta)(w-gamma) = 0$$
When we expand it, the cubic term cancel out and we are left with the equation of
a conic
$$(alpha vw + beta uw + gamma uv) - (ubetagamma + valphagamma + walphabeta) + alphabetagamma = 0tag{*1}$$
To see what sort of conic is this, substitute $w = 1 - u - v$ into LHS and expand, we get

$${rm LHS} = -alpha v^2 + (gamma - alpha - beta ) uv - beta u^2 + ( text{ linear/constant terms in } u, v )$$

Notice under affine transform, ellipse get mapped into ellipse. The conic above is an ellipse in Cartesian coordinates if and only if it is an ellipse in barycentric coordinates. In terms of coefficients in equation $(*1)$, we need
$$begin{align}
& 4alphabeta - (gamma - alpha - beta)^2 stackrel{?}{>} 0\
iff & 2(alphabeta + betagamma + gammaalpha) - alpha^2 - beta^2 - gamma^2 stackrel{?}{>} 0\
iff & alpha(beta + gamma - alpha) + beta(alpha+gamma-beta) + gamma(alpha+beta-gamma) stackrel{?}{>} 0
end{align}
$$
Since $a ge b, c$, we have $alpha ge beta, gamma$. The $2^{nd}$ and $3^{th}$ terms are already non-negative. Let's look at the $1^{st}$ term, we find

$$begin{align}beta + gamma - alpha
&= frac{b}{2R+b} + frac{c}{2R+c} - frac{a}{2R+a}\
&= left(frac{b}{2R+b} - frac{b}{2R+a}right) + left(frac{c}{2R+c} - frac{c}{2R+a}right) + frac{b+c-a}{2R+a}\
&= underbrace{frac{b(a-b)}{(2R+a)(2R+b)}}_{ge 0}
+ underbrace{frac{c(a-c)}{(2R+a)(2R+c)}}_{ge 0}
+ underbrace{frac{b+c-a}{2R+a}}_{> 0, text{ by triangle inequality}}\
&> 0
end{align}
$$
Combine all these, we can deduce $4alphabeta - (gamma - alpha - beta)^2 > 0$.
As a result, the conic in $(*1)$ is an ellipse in barycentric coordinates and hence one in Cartesian coordinates.

A Side Note

Let $J$ be the center of above ellipse. I was originally drawn to this question
by another question (currently deleted and OP decided to leave math.SE). Aside from labeling of the points, it is a conjecture about the areas of three triangles:

$$verb/Area/(IHJ) = verb/Area/(GFJ) = verb/Area/(EDJ)tag{*2}$$

In barycentric coordinates, we have

$$D = ( alpha, 0, 1-alpha), E = (0,beta,1-beta), F = (0,1-gamma,gamma),\
G = ( alpha,1-alpha,0), H = (1-beta,beta,0), I = (1-gamma,0,gamma)$$
For any $P = (u,v,w)$, it is not hard to show

$$begin{align}
verb/Area/(IHP)
&= Delta (v w - (v - beta)(w - gamma))
= Delta frac{partial}{partial u}mathcal{Q}(u,v,w)\
verb/Area/(GFP)
&= Delta( u w - (u - alpha)(w - gamma))
= Delta frac{partial}{partial v}mathcal{Q}(u,v,w)\
verb/Area/(EDP)
&= Delta( u v - (u - alpha)(v - beta))
= Delta frac{partial}{partial w}mathcal{Q}(u,v,w)
end{align}
$$
The conjecture about areas is equivalent to $J = (u_J,v_J,w_J)$ is a solution to the set of equations:
$$
frac{partial}{partial u}mathcal{Q}(u,v,w)
= frac{partial}{partial v}mathcal{Q}(u,v,w)
= frac{partial}{partial w}mathcal{Q}(u,v,w)tag{*3}$$

In Cartesian coordinates $(x,y)$, any conic can be viewed as the zero-set
of a quadratic polynomial $mathcal{P}(x,y)$. For circle/ellipse/hyperbola, the symmetric center is the unique point which satisfies

$$frac{partial}{partial x}mathcal{P}(x,y) = frac{partial}{partial y}mathcal{P}(x,y) = 0$$

Translate this to barycentric coordinates, the symmetric center of conic $mathcal{Q}(u,v,w) = 0$ is the unique point which satisfies $(*3)$. This means $J$ is a solution of $(*3)$ and hence conjecture $(*2)$ is true.