# 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 (HN = NO), while G is twice as close to O as it is to H (2·GO = HG).

 (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 HA, NA, GA, OA denote, respectively, the orthocenter, the 9-point center, the centroid, and the circumcenter of ΔBCD (A omitted!) Similarly introduce HB, NB, ..., and OD.

For a cyclic quadrilateral ABCD, all points OA- OD coincide and the quadrilaterals HAHBHCHD, NANBNCND, and GAGBGCGD 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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Let H' be the intersection of HAHC and HBHD, N' the intersection of NANC and NBND, G' the intersection of GAGC and GBGD, and O' the intersection of OAOC and OBOD. 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.

Discussion

### Reference

1. A. Myakishev, On Two Remarkable Lines Related to a Quadrilateral, Forum Geometricorum, Volume 6 (2006) 289-295. ### Discussion

The intersection of G' = GAGC ∩ GBGD is a natural construct that serves as the center of gravity, or centroid, of the quadrilateral ABCD. The other points

H' = HAHC ∩ HBHD,
N' = NANC ∩ NBND,
O' = OAOC ∩ OBOD

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.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 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, H'G' : G'O' = 2:1.

### Proof

Consider three affine maps fG, fO and fH transforming ΔABC onto triangles GAGBGC, OAOBOC, and HAHBHC, respectively. In the affine plane, write D = xA + yB + zC with x + y + z = 1, so that x, y, z are the barycentric coordinates of D with respect to ΔABC.

First note that

 fG(D) = fG(xA + yB + zC) = x·fG(A) + y·fG(B) + z·fG(C) = x·GA + y·GB + z·GC = 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 = GD.

Further, triangles ABC and OAOBOC are orthologic with centers D and OD. It is known [Danneels and Dergiades, Theorem 1] that the centers have the same barycentric coordinates relative to the respective triangles. Therefore, fO(D) = OD.

Also, since HXGX : GXOX = 2:1, for X = A, B, C, it follows that

fH(X)fG(X) : fG(X)fO(X) = 2:1,

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

fH(D)fG(D) : fG(D)fO(D) = 2:1,

or,

fH(D)GD : GDOD = 2:1

so that fH(D) = HD.

Finally, if Q = AC ∩ BD, then

fH(Q) = HAHC ∩ HBHD = H',
fG(Q) = GAGC ∩ GBGD = G',
fO(Q) = OAOC ∩ OBOD = O',

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

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

### Reference

1. E. Danneels and N. Dergiades, A theorem on orthology centers, Forum Geom, 4 (2004) 135-141. 