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Given any triangle $\triangle ABC$, let consider two sides, e.g. $AC$ and $BC$.

We draw two ellipses, one with foci in $A,C$ passing by $B$, and the other one with foci in $B,C$ and passing by $A$, obtaining two points $D$ and $E$ where the two ellipses intersect.

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Now, we draw two straight lines, one passing by $A$ and $C$ and the other passing $B$, $C$, annotating the intersection points $F,G,H,I,J,K$ with the two ellipses.

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My conjecture is that

Given any triangle $\triangle ABC$, the six points $F,H,D,G,I,E$ and the six points $A,J,D,K,B,E$ always determine two conic-sections.

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NOTE: This conjecture is inspired by this and this one, although I was not able so far to use the tools suggested therein to prove it.

Thanks for any suggestion for a compact proof!

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Your conjecture is true and can be proved as follows.

Consider the conics $\Gamma=DHFEBK$, $\Gamma'=DJAEIG$ and $\Psi=DHFEJ$. We have to prove that $G\in\Psi$.

Since $C$ is a common focus of $\Gamma$ and $\Gamma'$, they corrensponds under the homology $\Lambda$ of center $C$ and axis $DE$ sending $B$ into $J$. Consequently, we have $\Lambda(K)=G$.

In the conics $\Gamma$ and $\Psi$ consider, respectively, the hexagons $DEFKBH$ and $DEFG'JH$ where $G'$ is the intersection of $\Psi$ with $AC$ other than $F$. Then by Pascal's Theorem the lines $KB$ and $G'J$ meets on $DE$. Consequently, $G'$ must to be the image of $K$ under $\Lambda$, hence $$G'=\Lambda(K)=G$$ which concludes the proof.

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