To understand the argument consider first a circumscribed conic and the corresponding point E and line e, as described by the above reference. Then fix an equilateral triangle A'B'C' and its center E'. There is a unique projectivity H mapping A, B, C, E respectively to A', B', C', E'. Considering the harmonic tetrad (BCA*A'') and its image under H one sees that line e maps under H onto the line at infinity and A'' onto the middle of B'C' and the conic itself maps onto the circumcircle of A'B'C'. It follows that every conic circumscribing ABC can be obtained through the inverse projectivity F = H

A particular case represents the circumcircle of triangle ABC. The corresponding E being the symmedian point and e being the Lemoine axis of the triangle. Another case of interest is the Steiner circumellipse. To obtain this conic one identifies E with the centroid (intersection point of medians) of ABC.

By considering the composition G=F'*F

CrossRatio0.html

CrossRatioLines.html

GoodParametrization.html

Harmonic.html

Harmonic_Bundle.html

HomographicRelation.html

HomographicRelationExample.html

Pascal.html

PascalOnTriangles.html

ProjectivityFixingVertices.html

Steiner_Ellipse.html

TriangleCircumconics.html

TriangleConics.html

TriangleCircumconics2.html

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