An ellipse intrinsically bound to any triangle Given any triangle $\triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$. The hyperbole always intersects the side of the triangle that is opposite to the vertex through which it pass in two points $D$ and $E$.

Similarly, we can build other two hyperboles, one with foci in $A$ and $C$ and passing through $B$ (red), and one with foci in $B$ and $C$ and passing through $A$ (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.

Showing that the points in question lie on a common conic is straightforward.
I've renamed the points thusly: $D_B$ and $D_C$ are the points where the hyperbola through $A$ meets $\overline{BC}$; the subscripts indicate the closer vertex. Likewise for $E_C$, $E_A$, $F_A$, $F_B$.

Now, simply note that 
$$|BD_B| = |CD_C| \qquad |CE_C|=|AE_A| \qquad |AF_A| = |BF_B|$$
$$|D_BC| = |D_CB| \qquad |E_CA|=|E_AC| \qquad |F_AB| = |F_BA|$$
so that

$$\frac{BD_B}{D_BC}\cdot\frac{CE_C}{E_CA}\cdot\frac{AF_A}{F_AB} = 
\frac{CD_C}{D_CB}\cdot\frac{AE_A}{E_AC}\cdot\frac{BF_B}{FB_A} \tag{$\star$}$$

It happens that $(\star)$ holds if and only if $D_B$, $E_C$, $F_A$, $D_C$, $E_A$, $F_B$ lie on a conic. (This is the same condition used in this answer, except that it really doesn't matter here if we consider the ratios as signed or unsigned.) 
Showing that the conic is specifically an ellipse takes some extra work. I'll have to come back to that.
