Consider a triangle ABC and its Symmedian point K. The circle with diameter OK, O being the circumcenter of ABC is called Brocard circle of the triangle. The second intersection points of the symmedians AK, BK, CK build a triangle HEG called second Brocard triangle of ABC. Here are some elementary properties of these vertices:
 The symmedian through B passes through the common point D of tangents at A and C to the circumcircle (c).
 E is the common point of the symmedian AK, Brocard circle with diameter OK and circle (ADC).
 E is the middle of the chord AF of the circumcircle defined by the symmedian.
 angle(AEC) = 2*angle(B) and the symmedian BE is bisector of angle(AEC).
 Triangles BEA and CEB are similar.
Let (d) be the circle through A, C, D. Projecting on the sides and measuring the ratio DDC/DDB we find it equal to AB/BC, hence D on the symmedian AK. The circumcircle (d) of ACD passes through O, and intersects the Brocard circle at a point viewing KO under a right angle, hence coinciding with E. Since OE is orthogonal to the chord BF, E is its middle. These arguments proove [1, 2, 3].  is an easy consequence in view of the cyclic quadrangle AECD.
To prove  notice the equality of angles at E and that angle(BAE) = angle(A)-angle(EAC). But from triangle BCD angle(EBC) = angle(DCDB)-angle(EDC) = angle(A) - angle(EAC).
The vertices of the second Brocard triangle are the focal points of the Artzt parabolas (both kinds, first and second as well) of triangle ABC. To prove this one uses the characteristic property of E: (i) to be on the symmedian and (ii) to bisect angle AEC (see Artzt.html ).