QL-8P1: Steiner Angle Bisector Center Octet

EQF
In 1828 Jakob Steiner
published in Gergonne’s Annales 10 rules on the complete quadrilateral.
For a complete description see Ref-4.
Rules 8, 9, 10 are:
(8) Each of the four possible triangles has an incircle and three excircles.
The centers of these 16 circles lie, four by four, on eight new circles.
(9) These eight new circles form two sets of four, each circle of one set being
orthogonal to each circle of the other set. The centers of the circles of each
set lie on a same line. These two lines are perpendicular.
(10) Finally, these last two lines intersect at the point F (Miquel Point). Coordinates:

CT-coordinates:
In CT-notation let L1 = (1 : 0 : 0), L2 = (0 : 1 : 0), L3 = (0 : 0 : 1), L4 = (l : m : n).
Let Bij be the bisector (internal or external) between the lines Li and Lj.
Let Tj be intersection point Bij ^ Bjk, where i , j, k are consecutive numbers in the cycle (1,2,3,4).
Let T1, T2, T3, T4 be the points that define the Steiner Circles in rule (8).
The coordinates of these points are:
T1 = (–a                        : +b                      : +c     )
T2 = (+b m + c n - W : –b l                    : –c l   )
T3 = (–a m                   : +a l – c n + W : +c m)
T4 = (+a n                    : –b n                  : –a l + b m – W)
where W = [a2 (l - m) (l - n) + b2 (m - l) (m - n) + c2 (n - l) (n - m)].

Whether T1, T2, T3, T4 were created by “internal” or “external” bisectors (which actually is arbitrary in a quadrilateral) is only dependent on the signs in T1, T2, T3, T4 of the variables a, b, c, W. So the undetermined status of the bisectors being “internal” or “external” finds an algebraic solution in a combination of signs of a, b, c and W, as described here:
X1 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution a -> – a
X2 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution b -> – b
X3 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution c -> – c
X4 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution W -> –W
X1, X2, X3, X4 are collinear.
X5 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution a -> –a, W-> –W
X6 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution b -> –b, W-> –W
X7 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with substitution c -> –c, W-> –W
X8 = Perp. Bisector ( T1,T2) ^ Perp. Bisector ( T3,T4) with no substitution
X5, X6, X7, X8 are collinear.

Finally X1 – X8 end up in an algebraic expression that is not quite simple.
Here are the coordinates of X1 (with the substitution a -> a):
(a (a2 b l2 -b2 (a + c) m2 + c (a2 -b2 + 2 a c + b c + c2) n2 - (a + b) (a - b + c) n W - (a + c) W2 + m (b (a + b) (a -b + c) n
+ 2 b (a + c) W) + l (-a b (a - b + c) m + (-a2 b + b3 - a2 c + b2 c - 2 a c2 - b c2 - c3) n + a (a - b + c) W))  :
b (a (-b2 + a c + c2) l2 + b3 m2 + c (a2 - b2 + a c) n2 - (a + b) (a - b + c) n W + b W2 + m (b (a + b) (a - b + c) n
- 2 b2 W) + l (b (a - b + c) (b + c) m + (-a2 b + b3 - 2 a2 c - 2 a b c - 2 a c2 - b c2) n - (a - b + c) (b + c) W))  :
c (a (a2 + a b - b2 + 2 a c + c2) l2 - b2 (a + c) m2 + b c2 n2 +c (a - b + c) n W - (a + c) W2 + m (-b c (a - b + c) n
+ 2 b (a + c) W) + l (b (a - b + c) (b + c) m + (-a3 - a2 b + a b2 + b3 - 2 a2 c -a c2 - b c2) n - (a - b + c) (b + c) W)) )
The coordinates of X2 – X8 are similar.

Properties:
• M.V. Thébault described in Ref-39 that the ends of the diameters perpendicular to the corresponding line Li of the circumcircles of the component triangles Lj.Lk.Ll lie on two perpendicular axes through the Miquel Point. These axes happen to be the Steiner Axes with the 8 points QL-8P1. He also described that these Axes are the angle bisector of Vi.F.Vi’, where F=Miquel Point QL-P1, Vi=some vertex of some QL-Component Triangle and Vi’ = intersection point of 4th line with opposite line of Vi in QL-Component Triangle. See also QL-Tf1.
• The Centroid of the 8 centers of circles as described in rule (9) from Steiner (seen as a system of 8 random points) is QL-P4 (Miquel Circumcenter).
• The Steiner Axes (containing resp. X1,X2,X3,X4 and X5,X6,X7,X8) are the same axes as used in the construction of QL-Tf1 (Clawson-Schmidt Conjugate).
• Seiichi Kirikami describes another method for constructing QL-8P1: Let O1, O2, O3, O4 be the circumcenters of the Reference Quadrilateral. The in-ex-centers of the Component Triangles of O1.O2.O3.O4 lie, 4 by 4, on two orthogonal sets of 4 lines. Let La and Lb be the lines though QL-P1 parallel to these two sets. Denote the intersections of La and the lines of 1 set by A, B, C, D respectively. Denote the intersections of Lb and the lines of the other set by E, F, G, H respectively. The circles of their centers on A, B, C, D and radii A.QL-P1, B.QL-P1, C.QL-P1, D.QL-P1 intersect La at X1, X2, X3, X4 (4 of Steiner angle bisector center octet(QL-8P1)) respectively. The circles of their centers on E, F, G, H and radii E.QL-P1, F.QL-P1, G.QL-P1, H.QL-P1 intersect La at X5, X6, X7, X8 (4 of Steiner angle bisector center octet(QL-8P1)) respectively. See Ref-34, QFG # 1098.