Question

If three tangents to a circle form an equilateral triangle, prove that the tangent points form an equilateral triangle inscribed in the circle

Answer 1

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 $FED$ is an equilateral triangle according to the wording of the problem, we have that the angles $\widehat{AFB}=\widehat{BEC}=\widehat{CDA}=60$.
We also know that the circle in green is Inscribed in $FED$.

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 $(FO)$ 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 $\widehat{OFB}=30$.

Now let's work on the $OFB$ triangle.
Knowing that $\widehat{OBF}=90$ (because $(EF)$ is tangent to the circle at B and $OB$ 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 $\widehat{FOB}=180-90-30=60$ (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:
$\widehat{BOE}=\widehat{EOC}=\widehat{COD}=\widehat{DOA}=\widehat{AOF}=60 degrees$.
Knowing the above we notice that $\widehat{BOC}=\widehat{COA}=\widehat{AOB}=120degrees$.

We're at the last part of our proof here:
Now notice that $\widehat{BOA}$ subtends the same arc on the circle  that $\widehat{BCA}$.
According to the inscribed angle theorem, an angle $\theta$ inscribed in a circle is half of the central angle $2\theta$ that subtends the same arc on the circle. 
Therefore $\widehat{BCA}=\frac{\widehat{BOA}}{2}=60degrees$

We can prove in a similar fashion that:  $\widehat{CAB}=\widehat{ABC}=60degrees$ 
Therefore all the angles of the $ABC$ triangle have a measure of 60 degrees, we conclude then that  $ABC$ is equilateral.