I added a figure so you can guide yourself throughout the proof I'm about to write, so I recommend that you download the picture beforehand and have this window and the picture's window open. Alright, let's get started!
Assuming that [tex]FED[/tex] is an equilateral triangle according to the wording of the problem, we have that the angles [tex]\widehat{AFB}=\widehat{BEC}=\widehat{CDA}=60[/tex].
We also know that the circle in green is Inscribed in [tex]FED[/tex].
The following applies to every inscribed circle inside a triangle:
The center of the inscribed circle of a triangle is the intercept of all three angle bisectors of the triangle.
The above theorem implies that the line [tex](FO) [/tex] is an angle bisector because it goes through the vertex F and the center of the inscribed circle.
The previous statement implies that the angle [tex]\widehat{OFB}=30[/tex].
Now let's work on the [tex]OFB[/tex] triangle.
Knowing that [tex]\widehat{OBF}=90[/tex] (because [tex](EF)[/tex] is tangent to the circle at B and [tex]OB[/tex] is a radius of the circle. If you're lost here, remember that a tangent to a circle is always perpendicular to the radius of the circle.) we can then derive that [tex]\widehat{FOB}=180-90-30=60[/tex] (because the sum of the measures of all angles in a triangle is always equal to 180 degrees).
In the same way, we can prove also that:
[tex]\widehat{BOE}=\widehat{EOC}=\widehat{COD}=\widehat{DOA}=\widehat{AOF}=60 degrees[/tex].
Knowing the above we notice that [tex]\widehat{BOC}=\widehat{COA}=\widehat{AOB}=120degrees[/tex].
We're at the last part of our proof here:
Now notice that [tex]\widehat{BOA}[/tex] subtends the same arc on the circle that [tex]\widehat{BCA}[/tex].
According to the inscribed angle theorem, an angle [tex]\theta[/tex] inscribed in a circle is half of the central angle [tex]2\theta[/tex] that subtends the same arc on the circle.
Therefore [tex]\widehat{BCA}=\frac{\widehat{BOA}}{2}=60degrees[/tex]
We can prove in a similar fashion that: [tex]\widehat{CAB}=\widehat{ABC}=60degrees[/tex]
Therefore all the angles of the [tex]ABC[/tex] triangle have a measure of 60 degrees, we conclude then that [tex]ABC[/tex] is equilateral.
