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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.

enter image description here

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.

enter image description here

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!

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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 $B$.

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    $\begingroup$ Thanks, Fabio! That's what I was looking for! $\endgroup$ – user559615 Aug 26 '18 at 8:51

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