Central line (geometry)

The central line associated with the incenter X₁ = 1 : 1 : 1 (also denoted by I) is

$${\displaystyle x+y+z=0.}$$

This line is the antiorthic axis of △ABC.[6]

• The isogonal conjugate of the incenter of △ABC is the incenter itself. So the antiorthic axis, which is the central line associated with the incenter, is the axis of perspectivity of △ABC and its incentral triangle (the cevian triangle of the incenter of △ABC).
• The antiorthic axis of △ABC is the axis of perspectivity of △ABC and the excentral triangle △I₁I₂I₃ of △ABC.[7]
• The triangle whose sidelines are externally tangent to the excircles of △ABC is the extangents triangle of △ABC. △ABC and its extangents triangle are in perspective and the axis of perspectivity is the antiorthic axis of △ABC.

The trilinear coordinates of the centroid X₂ (also denoted by G) of △ABC are:

$${\displaystyle {\frac {1}{a}}:{\frac {1}{b}}:{\frac {1}{c}}}$$

So the central line associated with the centroid is the line whose trilinear equation is

$${\displaystyle {\frac {x}{a}}+{\frac {y}{b}}+{\frac {z}{c}}=0.}$$

This line is the Lemoine axis, also called the Lemoine line, of △ABC.

• The isogonal conjugate of the centroid X₂ is the symmedian point X₆ (also denoted by K) having trilinear coordinates a : b : c. So the Lemoine axis of △ABC is the trilinear polar of the symmedian point of △ABC.
• The tangential triangle of △ABC is the triangle △T_AT_BT_C formed by the tangents to the circumcircle of △ABC at its vertices. △ABC and its tangential triangle are in perspective and the axis of perspectivity is the Lemoine axis of △ABC.

The trilinear coordinates of the circumcenter X₃ (also denoted by O) of △ABC are:

$${\displaystyle \cos A:\cos B:\cos C}$$

So the central line associated with the circumcenter is the line whose trilinear equation is

$${\displaystyle x\cos A+y\cos B+z\cos C=0.}$$

This line is the orthic axis of △ABC.[8]

• The isogonal conjugate of the circumcenter X₆ is the orthocenter X₄ (also denoted by H) having trilinear coordinates sec A : sec B : sec C. So the orthic axis of △ABC is the trilinear polar of the orthocenter of △ABC. The orthic axis of △ABC is the axis of perspectivity of △ABC and its orthic triangle △H_AH_BH_C.

The trilinear coordinates of the orthocenter X₄ (also denoted by H) of △ABC are:

$${\displaystyle \sec A:\sec B:\sec C}$$

So the central line associated with the circumcenter is the line whose trilinear equation is

$${\displaystyle x\sec A+y\sec B+z\sec C=0.}$$

• The isogonal conjugate of the orthocenter of a triangle is the circumcenter of the triangle. So the central line associated with the orthocenter is the trilinear polar of the circumcenter.

The trilinear coordinates of the nine-point center X₅ (also denoted by N) of △ABC are:[9]

$${\displaystyle \cos(B-C):\cos(C-A):\cos(A-B).}$$

So the central line associated with the nine-point center is the line whose trilinear equation is

$${\displaystyle x\cos(B-C)+y\cos(C-A)+z\cos(A-B)=0.}$$

• The isogonal conjugate of the nine-point center of △ABC is the Kosnita point X₅₄ of △ABC.[10][11] So the central line associated with the nine-point center is the trilinear polar of the Kosnita point.
• The Kosnita point is constructed as follows. Let O be the circumcenter of △ABC. Let O_A, O_B, O_C be the circumcenters of the triangles △BOC, △COA, △AOB respectively. The lines AO_A, BO_B, CO_C are concurrent and the point of concurrence is the Kosnita point of △ABC. The name is due to J Rigby.[12]

The trilinear coordinates of the symmedian point X₆ (also denoted by K) of △ABC are:

$${\displaystyle a:b:c}$$

So the central line associated with the symmedian point is the line whose trilinear equation is

$${\displaystyle ax+by+cz=0.}$$

• This line is the line at infinity in the plane of △ABC.
• The isogonal conjugate of the symmedian point of △ABC is the centroid of △ABC. Hence the central line associated with the symmedian point is the trilinear polar of the centroid. This is the axis of perspectivity of the △ABC and its medial triangle.

The Euler line of △ABC is the line passing through the centroid, the circumcenter, the orthocenter and the nine-point center of △ABC. The trilinear equation of the Euler line is

$${\displaystyle x\sin 2A\sin(B-C)+y\sin 2B\sin(C-A)+z\sin 2C\sin(A-B)=0.}$$

This is the central line associated with the triangle center X₆₄₇.

The Nagel line of △ABC is the line passing through the centroid, the incenter, the Spieker center and the Nagel point of △ABC. The trilinear equation of the Nagel line is

$${\displaystyle xa(b-c)+yb(c-a)+zc(a-b)=0.}$$

This is the central line associated with the triangle center X₆₄₉.

The Brocard axis of △ABC is the line through the circumcenter and the symmedian point of △ABC. Its trilinear equation is

$${\displaystyle x\sin(B-C)+y\sin(C-A)+z\sin(A-B)=0.}$$

This is the central line associated with the triangle center X₅₂₃.