A regular hexagon in a triangle

Morley's famous trisector theorem delivered us equilateral triangles from a triangle ABC. There are more ways to create an equilateral triangle from ABC. But a regular hexagon ...

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The Fermat/Torricelli point or the 1st isogonic point

The starting point for the regular hexagon we'll be going to construct is the Fermat/Torricelli point. (See also my Dutch page isogonale driehoeken).

The Fermat/Torricelli point (F) can be constructed by attaching equilateral triangles on the outsides of the sides of ABC. Draw the segments from the outward pointed vertices of the equilateral triangles to the well chosen vertices of ABC, en the three segments concur in the point F. It is called the 1st isogonic point sometimes, because (when the angles of ABC are smaller than 120 degrees) angles AFB, BFC and CFA are all 120 degrees.

Six triangles inside ABC

We draw three lines parallel to the sides of ABC ("parallelians") passing through F:

In this way we get six similar triangles inside ABC: AKG, JBM, ILC, JKF, FLM and IFG. They form the base for the regular hexagon!

The regular hexagon!

The regular hexagon is found by taking the six Fermat/Torricelli points of the six similar triangles!

For further study read
Bernard Gibert and Floor van Lamoen, The Parasix Configuration and Orthocorrespondence, Forum Geometricorum 3 (2003) 169-180.

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