Given any triangle $\triangle ABC$, we can draw two ellipses, one with foci in $A,B$ and passing by $C$, and one with foci in $C,B$ and passing by $A$. We always obtain the points $D,E$, where these two ellipses intersect.
What I find nice and interesting (although it is likely an obvious property, sorry in this case!) is that
The ellipse with foci in $D,B$ and passing by $A$ pass also by $C$, whereas the one with foci in $E,B$ and passing by $C$ pass also by $A$,
as illustrated in the following picture.
To prove this, I tried to use the coordinates, but my calculations are too complicated, and I wonder if there is a more elementary way to show this result.
Thanks for your suggestions, and sorry again if this is too trivial!
The property can be proved by recalling the ellipse being the locus of points whose sum of the distances to the two focal points is constant.
Since $C$ and $D$ belongs to the ellipse with foci $A$ and $B$ and since $A$ and $D$ belongs to the ellipse with foci $C$ and $B$ we have: \begin{align} &\overline{AC}+\overline{CB}=\overline{AD}+\overline{DB}& &\overline{CA}+\overline{AB}=\overline{CD}+\overline{DB} \end{align} Subctrating we get $\overline{DA}-\overline{DC}=\overline{CB}-\overline{AB}$ that's $\overline{DA}+\overline{AB}=\overline{DC}+\overline{CB}$ which implies that $A$ and $C$ belongs to the ellipse with foci $D$ and $B$.