# Remarkable Line in a Quadrilateral

In a triangle, the orthocenter H, the 9-point center N, the centroid G, and the circumcenter O are collinear. The line that houses these four and other remarkable points is known as Euler's line of the triangle. The four points always occur on the Euler line in that order: H, N, G, O. Furthermore, N is always the midway between H and O

(1) |
HN : NO = 1:1 HG : GO = 2:1 |

In a quadrilateral ABCD, the four points A, B, C, and D form four triangles: ABC, BCD, CDA, and DAB, each with its own orthocenter, 9-point center, centroid, circumcenter, and Euler line. Let H_{A}, N_{A}, G_{A}, O_{A} denote, respectively, the orthocenter, the 9-point center, the centroid, and the circumcenter of ΔBCD (A omitted!) Similarly introduce H_{B}, N_{B}, ..., and O_{D}.

For a cyclic quadrilateral ABCD, all points _{A}- O_{D}_{A}H_{B}H_{C}H_{D}, N_{A}N_{B}N_{C}N_{D}, and G_{A}G_{B}G_{C}G_{D} happen to be homothetic with ABCD such that all three centers of homothety (denoted H, N, G) are collinear with the common circumcenter (O). The line of the centers served as a strong candidate for the title of *Euler line* of the quadrilateral ABCD. Points H, N, G, O lie on the line in that order but, instead of (1), they satisfy different proportions:

(2) |
HN : NO = 1:2 HG : GO = 1:1 |

It is perhaps fitting that towards the tricentennial anniversary of Euler's birth (15 April 1707) a new line was discovered that has even a stronger claim to the title of *Euler line* than the line of homothety centers. The applet below illustrates the concept.

What if applet does not run? |

Let H' be the intersection of H_{A}H_{C} and H_{B}H_{D}, N' the intersection of N_{A}N_{C} and N_{B}N_{D}, G' the intersection of G_{A}G_{C} and G_{B}G_{D}, and O' the intersection of O_{A}O_{C} and O_{B}O_{D}. Then in any quadrilateral, points H', N', G', O' are collinear and similar to (1),

(1') |
H'N' : N'O' = 1:1 and H'G' : G'O' = 2:1. |

### Reference

- A. Myakishev,
__On Two Remarkable Lines Related to a Quadrilateral__,*Forum Geometricorum*, Volume 6 (2006) 289-295.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

### Discussion

The intersection of G' = G_{A}G_{C} ∩ G_{B}G_{D} is a natural construct that serves as the center of gravity, or *centroid*, of the quadrilateral ABCD. The other points

H' = H_{A}H_{C} ∩ H_{B}H_{D},

N' = N_{A}N_{C} ∩ N_{B}N_{D},

O' = O_{A}O_{C} ∩ O_{B}O_{D}

by association, are called quasiorthocenter, quasi-nine-point-center, and quasicircumcenter, respectively. Quite obviously, the quasicircumcenter lies at the intersection of the diagonals AC and BD such that, for a cyclic quadrilateral, O' is indeed the circumcenter.

What if applet does not run? |

A. Myakishev credits Jaroslav Ganin with the statement of the theorem below and François Rideau with the idea of the proof.

### Theorem

In any arbitrary quadrilateral the quasiorthocenter H', the centroid G', and the quasicircumcenter O' are collinear. Furthermore,

### Proof

Consider three affine maps f_{G}, f_{O} and f_{H} transforming ΔABC
onto triangles G_{A}G_{B}G_{C}, O_{A}O_{B}O_{C}, and H_{A}H_{B}H_{C}, respectively. In the affine plane, write

First note that

f_{G}(D) | = f_{G}(xA + yB + zC) |

= x·f_{G}(A) + y·f_{G}(B) + z·f_{G}(C) | |

= x·G_{A} + y·G_{B} + z·G_{C} | |

= x·(B + C + D)/3 + y·(A + C + D)/3 + z·(A + B + D)/3 | |

= [(y + z)·A + (x + z)·B + (x + y)·C + (x + y + z)·D]/3 | |

= [(y + z)·A + (x + z)·B + (x + y)·C + (xA + yB + zC) | |

= [(x + y + z)·(A + B + C)]/3 | |

= (A + B + C)/3 | |

= G_{D}. |

Further, triangles ABC and O_{A}O_{B}O_{C} are orthologic with centers D and O_{D}. It is known [Danneels and Dergiades, Theorem 1] that the centers have the same barycentric coordinates relative to the respective triangles. Therefore, _{O}(D) = O_{D}.

Also, since H_{X}G_{X} : G_{X}O_{X} = 2:1, for X = A, B, C, it follows that

f_{H}(X)f_{G}(X) : f_{G}(X)f_{O}(X) = 2:1,

for X = A, B, C, and thus for any X in the plane. In particular,

_{H}(D)f_{G}(D) : f_{G}(D)f_{O}(D) = 2:1,

or,

f_{H}(D)G_{D} : G_{D}O_{D} = 2:1

so that f_{H}(D) = H_{D}.

Finally, if Q = AC ∩ BD, then

f_{H}(Q) = H_{A}H_{C} ∩ H_{B}H_{D} = H',

f_{G}(Q) = G_{A}G_{C} ∩ G_{B}G_{D} = G',

f_{O}(Q) = O_{A}O_{C} ∩ O_{B}O_{D} = O',

from which we conclude that H'G' : G'O' = 2:1.

The proportion H'N' : N'O' = 1:1 is treated similarly.

### Reference

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

68355383