Let for the areas $|\Delta ABP| =: F_B$, $|\Delta PCQ| =: F_C$, $|\Delta AQD| =: F_D$ and $|\Delta APQ| =: F_\Delta$.
I know $~~~ F_\Delta = \sqrt{(F_B+F_C+F_D)^2-4 F_B F_D}~~~$ (see here);
is there an easy way to show, that $$F_B+F_D=F_C,~~$$ if $\Delta APQ$ is an equilateral triangle
with that formula above for $F_\Delta$?
Just as example for an equilateral triangle:
To proved the equality $F_B + F_D = F_C$ when $\triangle APQ$ is equilateral, you don't need that formula of $F_\Delta$. It can be proved directly using a little bit of trigonometry.
When $\triangle APQ$ is equilateral, let $s$ be its side and let $\alpha, \beta, \gamma$ be angles illustrated below.
The three angles can be parameterized by a single $\theta \in (-\frac{\pi}{12},\frac{\pi}{12})$ as
$$(\alpha,\beta,\gamma) = \left(\frac{\pi}{12} + \theta, \frac{\pi}{12} - \theta, \frac{\pi}{4} - \theta\right)$$
In terms of these angles, the areas of the three right angled triangles are $$\begin{align} F_B &= \frac12 (s\cos\alpha)(s\sin\alpha) = \frac{s^2}{4}\sin(2\alpha) = \frac{s^2}{4} \sin\left(\frac{\pi}{6} + 2\theta\right)\\ F_D &= \frac12 (s\cos\beta)(s\sin\beta) = \frac{s^2}{4}\sin(2\beta) = \frac{s^2}{4} \sin\left(\frac{\pi}{6} - 2\theta\right)\\ F_C &= \frac12 (s\cos\gamma)(s\sin\gamma) = \frac{s^2}{4}\sin(2\gamma) = \frac{s^2}{4}\sin\left(\frac{\pi}{2} - 2\theta\right) = \frac{s^2}{4}\cos(2\theta)\end{align}$$
With help of the trigonometric identity $$\sin(\phi+\psi) + \sin(\phi-\psi) = 2\sin \phi\cos \psi$$ and the fact $\sin\frac{\pi}{6} = \frac12$, we find
$$F_B + F_D = \frac{s^2}{4}\times 2 \sin\frac{\pi}{6} \cos(2\theta) = \frac{s^2}{4}\cos(2\theta) = F_C$$