Thomson cubic (K002)
Here we make the same construction as that for K001. We construct a triangle and then the Thombson cubic for this triangle. Then we take any point on that cubic, after that we cojugate this point isogonal and we get another point which is on the same cubic. By using GeoGebra we conclude that Thombson cubic is isogonal transform of itself.
Barycentric equation
Proof
Here we again substitude , and with , and .
And again we have the same equation as that in the beginning. Thomson cubic is isogonal transform of itself.