EQUISEGMENTARY LINES

. In this paper we introduce the concept of equisegmentary lines in the isotropic plane. We derive the equations of equisegmentary lines for a standard triangle and prove that the angle between them is equal to the Brocard angle of a standard triangle. We study the dual Brocard circle, the circle whose tangents are equisegmentary lines, as well as the inertial axis and the Steiner axis. Some interesting properties of this circle are also investigated.


Introduction
The isotropic plane is a projective-metric plane, where the absolute consists of a line, the absolute line ω, and a point on that line, the absolute point Ω.Lines through the point Ω are isotropic lines and points on the line ω are isotropic points.
The distance between two points P i = (x i , y i ) (i = 1, 2) in the isotropic plane is defined by d(P 1 , P 2 ) = x 2 −x 1 and if x 1 = x 2 we say that P 1 and P 2 are parallel.For two parallel points P 1 , P 2 we define their span by s(P 1 , P 2 ) = y 2 − y 1 .The angle of two lines with equations y = k i x + l i (i = 1, 2) is k 2 − k 1 and if k 1 = k 2 we say that they are parallel.Any isotropic line is perpendicular to any non-isotropic line.Facts about the isotropic plane can be found in [11,12].
We say that a triangle is allowable if none of its sides is isotropic.If we choose the coordinate system in such a way the circumscribed circle of an allowable triangle ABC has the equation y = x 2 and therefore its vertices are the points A = (a, a 2 ), B = (b, b 2 ), and C = (c, c 2 ), while a + b + c = 0, we say that the triangle ABC is in standard position or shorter triangle ABC is a standard triangle.Its sides BC, CA, and AB have equations y = −ax − bc, y = −bx − ca, and y = −cx − ab.In order to prove geometric facts for any allowable triangle, it suffices to prove it for a standard triangle [7].
Denoting p = abc and q = bc + ca + ab, the authors proved a number of useful equalities in [7], e.g.
We proved also the following identities In the isotropic plane we have the following formula for Brocard angle of standard triangle:

Equisegmentary lines of a triangle
The motivation for this consideration are two Ocagne's papers [9, p. 131], and [10, p. 265].In this section we consider the equisegmentary lines of a standard triangle ABC.According to [7], standard triangle ABC has the centroid G = 0, − 2 3 q and the inertial axis with the equation y = − 2 3 q.Theorem 2.1.Let D 1 , E 1 , F 1 and D 2 , E 2 , F 2 be points on the lines BC, CA, and AB such that For variable u the centroids G 1 and G 2 of the triangles D 1 E 1 F 1 and D 2 E 2 F 2 lie on the inertial axis of the triangle ABC.The points G 1 and G 2 are symmetric with respect to the centroid G of that triangle (Figure 1).
which lie on the inertial axis with equation y = − 2 3 q and its midpoint is the point F 2 be points on the lines BC, CA, and AB such that where ω is the Brocard angle of the triangle ABC.The points D 1 , E 1 , F 1 and D 2 , E 2 , F 2 respectively, lie on one of the lines P 1 and P 2 .If ABC is a standard triangle, then We say that lines P 1 and P 2 from Corollary 2.3 are equisegmentary lines of the triangle ABC.These lines are reciprocal with respect to the triangle ABC.
In [9] and [10] d'Ocagne considered the equisegmentary lines in Euclidean geometry and obtained two pairs of reciprocal equisegmentary lines.
Theorem 2.4.Equisegmentary lines P 1 and P 2 of a standard triangle ABC have equations Proof.From (2.3) and (2.4) for the slopes of lines and respectively, and analogously lines and respectively.After adding and dividing by 3 we get the slopes k 1 and k 2 of lines P 1 and P 2 It remains to prove that lines P 1 and P 2 with equations (2.6a) and (2.6b) pass through the points G 1 and G 2 from (2.5).For the point G 1 and line P 1 we get The proof for the point G 2 and line P 2 is obtained by switching the indices 1 and 2. Theorem 2.5.The intersection of lines P 1 and P 2 from Theorem 2.4 is the point and the angle between P 1 and P 2 is equal to the Brocard angle of the triangle ABC (Figure 3).
Proof.The point P from (2.7) lies on the line P 1 with equation (2.6a) because and similarly for the line P 2 .
The angle of lines P 1 and P 2 equals Theorem 2.6.Using the notation from Corollary 2.3 the lines and determine the triangle with vertices 3) given by q , (2.9) Proof.The points E 1 and F 2 from (2.3) and (2.4) lie on the line and this is the line E 1 F 2 , and analogously for lines and by (2.8b) this x gives and this intersection is the point D ′ 1 from (2.9).
Theorem 2.7.The triangle D ′ 1 E ′ 1 F ′ 1 from Theorem 2.6 is homologous to the triangle ABC.The center of homology is the point and the axis of homology is the line P ′ 1 with equation Proof.The line with equation passes through the point A = (a, a 2 ) and the point D ′ 1 from (2.9) because and it is the line AD ′ 1 .This line passes also through the point P ′ 1 from (2.10) because and the same is also valid for lines BE ′ 1 and CF ′ 1 .For the abscissa of the intersection of lines E 1 F 2 , with equation (2.8a), and BC, given by y = −ax − bc, we get the equation with the solution and for the abscissa of the intersection of the line BC with the line P ′ 1 given by (2.11) we get the equation The equality of these two abscissae means that the point E 1 F 2 ∩ BC lies on the line P ′ 1 , and the same is also valid for points Visualization of the statements of Theorems 2.6 and 2.7.
According to [5] the bisectors of angles A, B, and C determine the triangle A s B s C s , the so called symmetral triangle of the triangle ABC, with vertices and by [4] the triangle ABC has the symmedian center K = 3p 2q , − q 3 .Comparing with the equalities (2.9) we directly get 6 is symmetrical to the symmetral triangle A s B s C s of the triangle ABC with respect to its symmedian center K (Figure 4). 2 , E ′ 2 , and F ′ 2 , which lie on the line P ′ 2 (Figure 5) given by (2.12) This implies y + ax = −bc, so this point lies on the line BC, i.e. we have However, this point also lies on the line P ′ 2 from (2.12) because The same is also valid for points The equality (a Theorem 2.11.Equisegmentary lines and the inertial axis of an allowable triangle ABC, and its Steiner axis touch a circle (Figure 6), which in the case of a standard triangle ABC has the equation Proof.According to [7], the inertial axis has the equation y = − 2 3 q, and this equation and (2.13) imply 1 8 x 2 − 6p q x + 9p 2 q 2 = 0 with double solution x = 3p q , and inertial axis touches the circle (2.13) at the point G ′ = 3p q , − 2 3 q .By [14] the Steiner axis S has the equation and from (2.13) and (2.14) we get the equation 1 8 x 2 + 6p q x + 9p 2 q 2 = 0 with double solution x = − 3p q .For this x, from (2.14) we obtain y = 9p 2 2q 2 − 2 3 q.Therefore the line S touches the circle (2.13) at the point Equation (2.6a) of the equisegmentary line P 1 and equation (2.13) give the following equation for the abscissa x which, because of and it is the point which lies on the same isotropic line with points Γ, S, and G ′′ .We just proved: Theorem 2.12.The circle from Theorem 2.11 touches the inertial axis of triangle ABC at the point symmetrical to its centroid with respect to the intersection of this inertial axis with the Brocard diameter of triangle ABC.This circle touches Steiner axis of this triangle at the midpoint of its Gergonne and its Steiner point.The isotropic line through the last three points is the bisector of the points of tangency of the considered circle with the equisegmentary lines of the triangle ABC (Figure 6).
Proof.Points D 1 and D 2 have abscissae c − u and b + u respectively.As they lie on the line BC with the equation y = −ax − bc, their ordinates are −a(c − u) − bc = c 2 + au, −a(b + u) − bc = b 2 − au.Because of that we have

Figure 2 .
Figure 2. Equisegmentary lines of the triangle ABC.

F 1 D
2 and D 1 E 2 .For the abscissa x of the intersection F 1 D 2 and D 1 E 2 , from (2.8b) and (2.8c), after multiplication by 2, we obtain the equation (b−c)x+b 2 −c 2 +ω(b+c−2a) = 0 with the solution

Figure 5 .
Figure 5. Visualization of the statement of Theorem 2.9.

2 Figure 6 .
Figure 6.Centroid G, Steiner point S, Gergonne point Γ, inertial axis G, Steiner axis S, equisegmentary lines P 1 and P 2 , Brocard diameter B, circumscribed circle K c , inscribed circle K i and the dual Brocard circle K ′ b of the triangle ABC (Visualization of the statements of Theorems 2.11, 2.12 and 2.13.)

2 from Theorem 2.1, lie on one of the lines P 1 and P 2 if and only if u = ω, where ω is the Brocard angle of the triangle ABC (Figure 2).
Visualization of the statement of Theorem 2.1.The points D 1 , E 1 , F 1 and respectively D 2 , E 2 , F