A conjecture related to any triangle

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 $$\frac12|\triangle ABC|\left(\;1+3(de+ef+fd)\;\right) = -|\triangle ABC|\qquad\text{when exactly two of $d$, $e$, $f$ match}$$

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As a side note, this recent question asks about the concurrency of segments through the apex vertices of the isosceles triangles in general.