The fact there are 6 congruent circls. These circles congruent to two Johnson circles and two Steiner circles. So there are 10 congruent circles in this conffiguration. Dao Thanh Oai- VietNam

@ToniMhax Can you help me show detail?

@FedorPetrov You can click here geogebra.org/m/hadhw364

I checked by geogebra, I think $AD, BE, CF$ concurrent if and only if the above trigonometric formula is equal to 1.

@LSpice I updated new Figure

Please help me check again. Because, I think with six lines $a, b, c, d, f$ above, there are only (there are maximum) $3$ Brianchon points $M, N, P$.

Your reference here but I don’t think it is the answer for second question.

About Pascal line: The 60 Pascal lines intersect three at a time through 20 Steiner points (some of which are shown as the filled circles in the above figures). In the symmetrical case of the regular hexagon inscribed in a circle, the 20 Steiner points degenerate into seven distinct points arranged at the vertices and center of a regular hexagon centered at the origin of the circle. The 60 Pascal line also intersect three at a time in 60 Kirkman points. Each Steiner point lines together….. See in here

How can proof these points are collinear?

Thank You I will correct title