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 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 O_A- O_D coincide and the quadrilaterals H_AH_BH_CH_D, N_AN_BN_CN_D, and G_AG_BG_CG_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.

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 H_AH_C and H_BH_D, N' the intersection of N_AN_C and N_BN_D, G' the intersection of G_AG_C and G_BG_D, and O' the intersection of O_AO_C and O_BO_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.

Discussion

Reference

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

Discussion

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

H' = H_AH_C ∩ H_BH_D,
N' = N_AN_C ∩ N_BN_D,
O' = O_AO_C ∩ O_BO_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, H'G' : G'O' = 2:1.

Proof

Consider three affine maps f_G, f_O and f_H transforming ΔABC onto triangles G_AG_BG_C, O_AO_BO_C, and H_AH_BH_C, 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

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_AO_BO_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, f_O(D) = O_D.

Also, since H_XG_X : G_XO_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,

f_H(D)f_G(D) : f_G(D)f_O(D) = 2:1,

or,

f_H(D)G_D : G_DO_D = 2:1

so that f_H(D) = H_D.

Finally, if Q = AC ∩ BD, then

f_H(Q) = H_AH_C ∩ H_BH_D = H',
f_G(Q) = G_AG_C ∩ G_BG_D = G',
f_O(Q) = O_AO_C ∩ O_BO_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

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

71930552

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.