Can you prove the following claim:
Claim. Given an arbitrary $\triangle ABC$. The isosceles $\triangle ADB$ , $\triangle BEC$ and $\triangle CFA$ are constructed on the sides of the $\triangle ABC$ externally . Let $F_1$,$F_2$,$F_3$ be the Fermat points of the $\triangle ADB$ , $\triangle BEC$ , $\triangle CFA$ , respectively. Then the $\triangle F_1F_2F_3$ is an equilateral triangle.
GeoGebra applet that demonstrates this claim can be found here.
Using a little bit of trigonometry one can show that the Fermat point lies at an angle of exactly $30^{\circ}$ relative to the base for all isosceles triangles , which means that $\angle BAF_1=\angle F_1BA=\angle CBF_2=\angle F_2CB=\angle ACF_3=\angle F_3AC = 30^{\circ}$ . I am stuck here. I don't know how to proceed with the proof. Any hints are welcomed.