A conjecture related to any triangle 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!
 A: 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,
$$|\triangle D_{d}E_{e}F_{f}| + |\triangle D_{-d}E_{-e}F_{-f}| = \frac12 |\triangle ABC| \left(\; 1 +t^2 (de+ef+fd) \;\right) \tag{$\star$}$$

For $\theta = 30^\circ$ (Napoleon), $t^2=1/3$, and $(\star)$ becomes
  $$\frac16|\triangle ABC|\left(\;3+de+ef+fd\;\right) = |\triangle ABC| \qquad\text{when $d$, $e$, $f$ match }$$
  And, for $\theta = 60^\circ$ (OP), $t^2 = 3$, so
  $$\frac12|\triangle ABC|\left(\;1+3(de+ef+fd)\;\right) = -|\triangle ABC|\qquad\text{when exactly two of $d$, $e$, $f$ match}$$ 


As a side note, this recent question asks about the concurrency of segments through the apex vertices of the isosceles triangles in general.
A: I guess you refer to oriented areas for the triangles.
I have built a geogebra model and moving B I see some cases that can match your conjecture only using this definition of area.
So this is a case similar to your first one.

If I move point B to reach segment AC, the triangle ABC degenerates to a segment (Area of ABC equals 0) but the other triangles do not.
Though, if B is any point of the straight line AB, both triangles IDG and EFH are symmetrical, so if we consider an oriented area, one has positive area, the other one has the same negative area and only in this case the sum is correct.
If I understand well this is a counter example to your conjecture in the original form.


