A simple conjecture (and a question) about three parabolas related to any triangle Given any triangle, we can build three parabolas, each with focus on one vertex and with directrix the opposing side, as illustrated here:

My first conjecture, likely trivial, is that, given any triangle, 

The three parabolas never intersect, but they are tangent to one another in at most three points. 

For instance, in case of an equilateral triangle, it seems that the three parabolas "touch" each other in three points $E,F,G$ 

However, it is not obvious to me whether the equilateral triangle is the only case in which we can find three tangential points $E,F,G$. 
The question is, then:

In which conditions (on the initial triangle) can we find three, two, one or no tangential points?

I apologize in case the question is trivial. But thank you very much for your hints, comments, suggestions!
 A: 
Let $\ell$ be the perpendicular bisector of $\overline{BC}$, and let $B^\prime$ be the point where $\ell$ meets the perpendicular at $B$ to $\overline{AB}$. Then $B^\prime$ is equidistant from $C$ and $\overleftrightarrow{AB}$, so $B^\prime$ is on the $C$-focus parabola. Since $\ell$ bisects $\angle BB^\prime C$, an aspect of the parabolic reflection property implies that $\ell$ is tangent to the parabola.
Similarly, $\ell$ is tangent to the $B$-focus parabola (say, at $C^\prime$), so $\ell$ separates the two parabolas, making the only possibility of their meeting when $B^\prime$ and $C^\prime$ coincide. This happens if and only if $\triangle ABC$ is isosceles with base $\overline{BC}$. $\square$
From this, we see that the three parabolas admit no tangential points iff the triangle is scalene, three tangential points iff the triangle is equilateral, and exactly one tangential point iff the triangle is non-equilaterally isosceles. There are never exactly two tangential points. $\square$
