[1] The symmedian through B passes through the common point D of tangents at A and C to the circumcircle (c).

[2] E is the common point of the symmedian AK, Brocard circle with diameter OK and circle (ADC).

[3] E is the middle of the chord AF of the circumcircle defined by the symmedian.

[4] angle(AEC) = 2*angle(B) and the symmedian BE is bisector of angle(AEC).

[5] Triangles BEA and CEB are similar.

Let (d) be the circle through A, C, D. Projecting on the sides and measuring the ratio DD

To prove [5] notice the equality of angles at E and that angle(BAE) = angle(A)-angle(EAC). But from triangle BCD angle(EBC) = angle(DCD

The vertices of the second Brocard triangle are the focal points of the

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