# Locus of a vertex defining a circle intrinsic to a triangle

Given a triangle $$\triangle ABC$$, we draw the circles with centers in two vertices (say, $$A,C$$ in the picture below) and passing through the third one (say, $$B$$), determining the points $$D$$ and $$E$$ on the side $$\overline{AC}$$.

Then, we can draw the two circles with centers in the same two vertices and passing by the $$D$$ and $$E$$, determining other two points $$F$$ and $$G$$ on the other two sides.

The five points $$B,D,E,F,G$$, as shown here, always determine a circle (red).

My question is:

Given two vertices of $$\triangle ABC$$, what is the locus determined by the third point in such a way that the red circumference is well defined?

My suspect is that this locus is the inner part of an ellipse (orange, in the picture below) passing through the two initial vertices, and two points $$H,I$$ located at the far vertices of two equilateral triangles built on the side defined by the two initial points, as illustrated here:

But I have difficulties to determine a fifth point $$J$$ which would define such an ellipse without any ambiguity.

Thanks for your help, and sorry in case of trivialities!

• Since the roles of the points with respect to the circle are different, you should specify which "two vertices of $\triangle ABC$" you intend to keep fixed. Also, to be clear: When you write "the red circumference is well defined", do you mean that the red circle is also fixed in place? – Blue Oct 31 '18 at 20:35
• @Blue You're right. I will edit it. Thanks! About the red circle, I mean that all the $5$ points $B,D,E,F,G$ (or are at least $3$ of them) are defined. Or, is there a a better way to phrase it? Sorry for confusion. – user559615 Oct 31 '18 at 20:51
• For $\triangle ABC$, pt $B$ is defined by fiat; $D$ and $E$ are always constructible on extended side $\overleftrightarrow{AC}$. Correspondingly, $F$ and $G$ are constructible on extended sides $\overleftrightarrow{AB}$ and $\overleftrightarrow{BC}$. Thus, there's always a circle through those five pts. I guess you want to confine $D$, $E$, $F$, $G$ to the non-extended segments. For $D$ and $E$ that's easy: just require $|AB|$ and $|BC|$ to be less than $|AC|$. (@Edward's football.) Then, $|AF|=|AD|=|AC|-|BC|<|AB|$ (by the Triangle Inequality), so $F$ is automatically okay; likewise, $G$. – Blue Oct 31 '18 at 21:38

According to GeoGebra, if $$A$$ and $$C$$ are the fixed vertices, the equilateral triangles are significant, but the figure within which circle $$BFDEG$$ is defined is not an ellipse but rather the figure $$HAIC$$ (reminiscent of Euclid Elements I, 1), that is, the football-shaped overlap of two circles with centers $$A$$, $$C$$ and intersections $$H$$, $$I$$.
Circle $$BFDEG$$ persists wherever $$B$$ is moved within that figure, but shrinks to a point when $$\triangle ABC$$ degenerates to a line (when $$B$$ lies on $$AC$$), and disappears altogether when $$B$$ is moved outside the figure.
addendum: As noted in response(s) to OP's July posting, a necessary condition of the five-point circle $$BFDEG$$, where $$A$$, $$C$$ are fixed, is that $$\angle BAC$$ and $$\angle BCA$$ are both acute. A further necessary condition seems to be that neither $$BA$$ nor $$BC$$ is greater than the given line $$AC$$. E.g. if $$AB>AC$$, then $$E$$ lies on $$AC$$ extended and $$GBDE$$ is no longer an isosceles trapezoid and/or inscribable in the same circle as isosceles trapezoid $$FBED$$ (cf. @dan fulea response to the July posting).
I note, however, that $$G'BDE$$ is an isosceles trapezoid concyclic with $$FBED$$, if $$G'$$ is diametrically opposite $$G$$ on the circle about $$C$$, i.e. on side $$BC$$ extended. But setting this aside as part of a more general problem, we seem to be left with the football.