Given one side $AC$ of any triangle $\triangle ABC$, we can draw the couple of circles with center in $A$ and passing through $C$ and with center in $C$ and passing through $A$, obtaining two points $D,E$ corresponding to the intersections of these two circles.
The same can be done with the other two sides, obtaining other $2$ couples of circles and other $4$ points $F,G,H,I$.
My conjecture is that
The sum of the areas of the triangles $\triangle IDG$ and $\triangle FHE$ is equal to the area of the triangle $\triangle ABC$.
(Notice that $I,D,G$ and $F,H,E$ are the intersection points of three distinct couples of circles, underlined by the different colors. I suspect that similar conjectures can be formulated starting from different choice of these points. E.g., if we consider the triangles $\triangle EHG$ and $\triangle FID$, the sum of the area of the first one and the area of the original triangle $\triangle ABC$ seems to give the area of the second one, as illustrated below).
This is probably a very obvious result, and I apologize in this case. However, I would need some hint to provide a compact proof of such conjecture. And I therefore thank you for any suggestion and comment!
Best Answer
This is a relative of a property of Napoleon Triangles, and/or what we might call "Petr-Douglas-Neuman" triangles: triangles determined by the apex vertices of isosceles triangles erected on the sides of a given triangle.
Let's coordinatize in the $xy$-plane of $\mathbb{R}^3$ (the third dimension will give us easy access to "signed area"), with $$A = (0,0,0) \qquad B = (c,0,0) \qquad C = (b \cos A, b \sin A,0)$$
Note that one traces $A\to B\to C\to A$ in a counterclockwise fashion.
Define $(x,y,0)^\perp := (-y,x,0)$, which performs a $90^\circ$ counterclockwise rotation of the vector $(x,y,0)$ about the origin. With this, we can define $$\begin{align} D_{\pm}&:=\frac12\left(\;(B+C)\mp t\;(C-B)^\perp\;\right) \\[4pt] E_{\pm}&:=\frac12\left(\;(C+A)\mp t\;(A-C)^\perp\;\right) \\[4pt] F_{\pm}&:=\frac12\left(\;(A+B)\mp t\;(B-A)^\perp\;\right) \end{align}$$ where $t := \tan\theta$ for $0<\theta<90^\circ$ the base angle of an isosceles triangle. Sign choices guarantee that $D_{+}$ is the apex of such an isosceles triangle erected upon $\overline{CA}$ outside of $\triangle ABC$, while $D_{-}$ is the apex of the isosceles triangle that overlaps $\triangle ABC$. (So, for Napoleon Triangles, $\theta = 30^\circ$, while for the triangles in this question, $\theta = 60^\circ$.)
We choose $+$ to be outside the triangle, so that $\triangle D_{+} E_{+} F_{+}$ is traced in the same orientation as $\triangle ABC$. (Note that $D_{-}E_{-}F_{-}$ may-or-may-not have the opposite orientation; it depends upon whether $\theta$ is big enough to cause the apices of the isosceles triangles to move past each other.)
Now, we calculate the signed area of $\triangle XYZ$ via $$|\triangle XYZ| := \left(\;(Y-X)\times(Z-X)\;\right)\cdot\left(0,0,\frac12\right)$$ Effectively, we take half the third coordinate of the cross product; the other two coordinates are $0$, anyway. (That's how the third dimension comes in handy!) With this definition, we have, for signs $d$, $e$, $f$, $$\begin{align} |\triangle ABC| &= \frac12b c \sin A \\[6pt] |\triangle D_d E_e F_f| &= \frac14 |\triangle ABC|\left(1+t^2(de+ef+fd)\right) + \frac18 \left(a^2 d + b^2 e + c^2 f\right) \\[6pt] |\triangle D_{-d} E_{-e} F_{-f}| &= \frac14 |\triangle ABC|\left(1+t^2(de+ef+fd)\right) - \frac18 \left(a^2 d + b^2 e + c^2 f\right) \end{align}$$ Thus,
$$|\triangle D_{d}E_{e}F_{f}| + |\triangle D_{-d}E_{-e}F_{-f}| = \frac12 |\triangle ABC| \left(\; 1 +t^2 (de+ef+fd) \;\right) \tag{$\star$}$$
As a side note, this recent question asks about the concurrency of segments through the apex vertices of the isosceles triangles in general.