In geometry, the Newton–Gauss line (or Gauss–Newton line) is the line joining the midpoints of the three diagonals of a complete quadrilateral. The midpoints of the two diagonals of a convex quadrilateral with at most two parallel sides are distinct and thus determine a line, the Newton line. If the sides of such a quadrilateral are extended to form a complete quadrangle, the diagonals of the quadrilateral remain diagonals of the complete quadrangle and the Newton line of the quadrilateral is the Newton–Gauss line of the complete quadrangle.
Complete quadrilaterals
Any four lines in general position (no two lines are parallel, and no three are concurrent) form a complete quadrilateral. This configuration consists of a total of six points, the intersection points of the four lines, with three points on each line and precisely two lines through each point. These six points can be split into pairs so that the line segments determined by any pair do not intersect any of the given four lines except at the endpoints. These three line segments are called diagonals of the complete quadrilateral.
Existence of the Newton−Gauss line
It is a well-known theorem that the three midpoints of the diagonals of a complete quadrilateral are collinear. There are several proofs of the result based on areas or wedge products or, as the following proof, on Menelaus's theorem, due to Hillyer and published in 1920. Let the complete quadrilateral ABCA'B'C' be labeled as in the diagram with diagonals AA', BB', CC' and their respective midpoints L, M, N. Let the midpoints of BC, CA', A'B be P, Q, R respectively. Using similar triangles it is seen that QR intersects AA' at L, RP intersects BB' at M and PQ intersects CC' at N. Again, similar triangles provide the following proportions,
R L ¯ L Q ¯ = B A ¯ A C ¯ , Q N ¯ N P ¯ = A ′ C ′ ¯ C ′ B ¯ , P M ¯ M R ¯ = C B ′ ¯ B ′ A ′ ¯ . {\displaystyle {\frac {\overline {RL}}{\overline {LQ}}}={\frac {\overline {BA}}{\overline {AC}}},\quad {\frac {\overline {QN}}{\overline {NP}}}={\frac {\overline {A'C'}}{\overline {C'B}}},\quad {\frac {\overline {PM}}{\overline {MR}}}={\frac {\overline {CB'}}{\overline {B'A'}}}.}
However, the line A'B'C intersects the sides of triangle △ABC, so by Menelaus's theorem the product of the terms on the right hand sides is −1. Thus, the product of the terms on the left hand sides is also −1 and again by Menelaus's theorem, the points L, M, N are collinear on the sides of triangle △PQR.
Applications to cyclic quadrilaterals The following are some results that use the Newton–Gauss line of complete quadrilaterals that are associated with cyclic quadrilaterals, based on the work of Barbu and Patrascu.
Equal angles
Given any cyclic quadrilateral ABCD, let point F be the point of intersection between the two diagonals AC and BD. Extend the diagonals AB and CD until they meet at the point of intersection, E. Let the midpoint of the segment EF be N, and let the midpoint of the segment BC be M (Figure 1).
Theorem If the midpoint of the line segment BF is P, the Newton–Gauss line of the complete quadrilateral ABCDEF and the line PM determine an angle ∠PMN equal to ∠EFD.
Proof First show that the triangles △NPM, △EDF are similar. Since BE ∥ PN and FC ∥ PM, we know ∠NPM = ∠EAC. Also, B E ¯ P N ¯ = F C ¯ P M ¯ = 2. {\displaystyle {\tfrac {\overline {BE}}{\overline {PN}}}={\tfrac {\overline {FC}}{\overline {PM}}}=2.}
In the cyclic quadrilateral ABCD, these equalities hold:
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