(i) An equilateral triangle A

(ii) A line (e) through the center of the equilateral,

(iii) A point I

A Moebius diagram results by defining the Moebius transformation F mapping the equilateral A

By such a transformation line (e) is the pre-image of the Brocard axis of ABC and I

Here I show that such a diagram uniquely characterises the similarity class of a triangle.

For this I fix the equilateral and its circumcircle c

There it was shown that F maps the inverse Q

The theorem results immediately from the fact that the equi-brocardian triangles inscribed in (c) result by the same Moebius transformation F, by applying it to an other equilateral A

The theorem follows as a corollary of the previous one, since similar triangles have the same Brocard angle and each similarity class is unique (up to rotations) isncribed in (c) among its equi-brocardians. But a direct proof is equally easy. In fact, if triangles ABC and A'B'C' are similar and the corresponding Moebius transformations are F and F', then

is a Moebius transformation between the triangles coinciding with the similarity on the vertices of ABC, which is also a Moebius transformation. Hence it coincides wholly with the similarity and we have

from which follows immediately that the Moebius diagram defined from F and F' coincide.

Inversely, if the Moebius diagrams coincide and {F, F'} are the corresponding Moebius transformations, then the transformation S defined by the previous formula is also a Moebius transformation fixing the point at infinity, since S(I) = F'(F

In fact, point I

To construct the Moebius transformation F, define an isosceles QWW' with angle at the apex twice the Brocard angle calculated above. Then define F as the unique Moebius transformation mapping the vertices of Q

In fact, denoting by {z,z'} the diametral intersections of c

Then, using the angle (fi) the vertex A can be located by drawing the circle passing through the isodynamic point J, orthogonal to c and making an angle (fi) with ww' and taking its intersection point with c. This is the Apollonian circle of the triangle ABC passing through A. The other Apollonian circles can be constructed similarly and define through their intersections with c the other vertices of the triangle ABC.

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