A conjecture about the intersections of three hyperboles related to any triangle

Question

Given any triangle $\triangle ABC$, we build the hyperbole with foci in $A$ and $B$ and passing through $C$.

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).

The first part of my conjecture is that the three hyperboles always intersect in two points $D$ and $E$.

Moreover, the ellipse with foci in these two points $D$ and $E$, and passing through one of the three vertices of the triangle $\triangle ABC$, pass also through the other two vertices.

These are probably obvious results. However, is there an elementary proof for these conjectures?

Thanks for your help! Sorry in case this is too trivial.

EDIT: You might be interested also in this other post.

Answer 1

Try this:

Let E be a point of intersection of black and red hyperbole. Then

black hyperbole: $EA-EB=AC-CB$

red hyperbole: $EA-EC=AB-BC$

Subtracting, you get

$$EC-EB=AC-AB$$ therefore green hyperbole also passes through E.

Do the same for point D, saying that D is a point of intersection of red and green hyperbole, and you will get that black hyperbole also passes through D.

Answer 2

Second part: Let $a=BC$, $b = CA$ and $c =AB$.

We have $$EB-EC = DC-DB = c-b$$ $$\color{red}{EB-EA = DA-DB = a-b}$$ $$DA-DC = FC-FA =c-a$$

If $$AD+AE = 2d$$ then since $$BD+BE = (DA+b-a)+(EA+a-b)=2d$$

so $B$ also lies on this ellipse. The same is true for $C$.