# An interesting (conjectural) property of any triangle

Given any triangle $$\triangle ABC$$, we can always build three ellipses, each of them having foci in two of the vertices and passing through the third one, as shown in the following picture:

In general, these three ellipses intersect in $$6$$ points $$D,E,F,G,H,I$$.

Given any two among these $$6$$ points, it is always possible to build an hyperbole with foci in the given points an passing by one vertex of the triangle $$\triangle ABC$$.

My conjecture is that

Any pair of these $$6$$ points represents the two foci of at least one hyperbole passing by at least two of the three vertices of $$\triangle ABC$$.

In this example, I choose the points $$D$$ and $$F$$ as foci. The hyperbole with foci in $$D$$ and $$E$$ and passing through $$A$$ (magenta), pass also through $$B$$.

Is it possible to prove this conjecture with a compact proof?

Thanks for your help! I apologize in case of mistakes, obscurity or trivialities.

In your picture, since $$D, F$$ are on the same ellipse with foci $$A$$ and $$B$$, so we have $$AD+BD = AF+BF$$. Hence we have $$AF - AD = BD - BF$$, which directly implies that the hyperbole with foci $$D$$ and $$F$$ and passing by $$A$$ also passes $$B$$.
• @AndreaPrunotto In your new picture, both are contained in the red ellipse. You can check that if you choose 2 points among $DEFGHI$, both are contained in one of the three ellipses. – Seewoo Lee Oct 18 '18 at 17:48
• I have notice that (still using my picture), the points $G$ and $E$ allow to build a hyperbole passing through all the three vertices of $\triangle ABC$. Any idea why these two points? – user559615 Oct 18 '18 at 18:01