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.

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.)