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

The complete quadilateral $ABCDEF (E = AD \cap BC, F = AB \cap CD)$ be given. Denote $O$ the midpoint $AC, O_1$ the midpoint $BD, O_2$ the midpoint $EF, H$ the orthocenter of $\triangle CDE, H_1$ the orthocenter of $\triangle BCF, H_2$ the orthocenter of $\triangle ABE, H_3$ the orthocenter of $\triangle ADF, \omega$ the circle with diameter $AC, \Omega$ the circle with diameter $BD, \theta$ the circle with diameter $EF, \alpha$ the circle with diameter $CD, \beta$ the circle with diameter $CE.$

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 $C_1 \in AD, CC_1 \perp DE, D_1 \in BC, DD_1 \perp CE, E_1 \in CD, EE_1 \perp CD \implies$ $H = CC_1 \cap DD_1 \cap EE_1, C_1 \in \alpha, D_1 \in \alpha, C_1 \in \beta, E_1 \in \alpha, \implies$ $H$ 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 and $HH_1$ 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.

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Gauss line

Existence of the Gauss line