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

User Jodonnell
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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.


If three tangents to a circle form an equilateral triangle, prove that the tangent-example-1
User J Lundberg
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