Gauss line

The Gauss line (or Gauss–Newton line) is the line joining the midpoints of the three diagonals of a complete quadrilateral.

Existence of the Gauss line

Gauss line 1.png

The complete quadilateral $ABCDEF$ $(E = AD \cap BC, F = AB \cap CD)$ be given. Denote $O, O_1, O_2$ the midpoints of $AC, BD, EF,$ respectively.

Denote $H, H_1, H_2, H_3$ the orthocenters of the $\triangle CDE, \triangle BCF, \triangle ABE, \triangle ADF,$ respectively.

Denote $\omega, \Omega,  \theta, \alpha,$ and $\beta$ the circles with diameters $AC, BD, EF, CD,$ and $CE,$ respectively.

Prove a) points $O, O_1, O_2$ are collinear;

b) $OO_1 \perp HH_1;$

c) points $H, H_1, H_2, H_3$ are collinear.

Proof

Let \[C_1 \in AD, CC_1 \perp AD, D_1 \in BC, DD_1 \perp BC, E_1 \in CD, EE_1 \perp CD \implies\] \[H = CC_1 \cap DD_1 \cap EE_1, C_1 \in \alpha, D_1 \in \alpha, C_1 \in \beta, E_1 \in \beta \implies\] $H$ is the radical center of $\omega, \Omega,$ and $\alpha \implies H$ lies on the radical axes of $\omega$ and $\Omega.$

$H$ is the radical center of $\omega, \theta,$ and $\beta \implies H$ lies on the radical axes of $\omega$ and $\theta.$

Similarly, using circles with diameters $BC$ and $FC$ one can prove that $H_1$ lies on the radical axes of $\omega$ and $\Omega$ and on the radical axes of $\omega$ and $\theta.$

Therefore $HH_1 \perp OO_1, HH_1 \perp OO_2 \implies$ points $O, O_1,$ and $O_2$ are collinear.

It is clear that $HH_1$ is the perpendicular to the line $OO_1O_2.$.

Similarly, one can prove that $H_2$ and $H_3$ lie on the radical axes of $\omega$ and $\Omega \implies$ points $H, H_1, H_2$ and $H_3$ are collinear.

vladimir.shelomovskii@gmail.com, vvsss