Newton–Gauss line

The line created by connecting the midpoints of all diagonals (L, M, and N) is the Newton–Gauss Line.

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.^[1] 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.^[2] There are several proofs of the result based on areas ^[2] or wedge products^[3] or, as the following proof, on Menelaus's theorem, due to Hillyer and published in 1920.^[4]

Let the complete quadrilateral ABCA'B'C' be labeled as in the diagram with diagonals AA' , BB' and CC' and their respective midpoints, L, M and N. Let the midpoints of BC, CA' and A'B be P, Q and 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,

${\displaystyle {\frac {RL}{LQ}}={\frac {BA}{AC}},\quad {\frac {QN}{NP}}={\frac {A'C'}{C'B}},\quad {\frac {PM}{MR}}={\frac {CB'}{B'A'}}.}$

However, the line AB'C' intersects the sides of triangle A'BC, 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 and 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.^[5]

Equal angles[]

Figure 1: An angle equality.

Given any cyclic quadrilateral ${\displaystyle ABCD}$, let point ${\displaystyle F}$ be the point of intersection between the two diagonals ${\displaystyle AC}$ and ${\displaystyle BD}$. Extend the diagonals ${\displaystyle AB}$ and ${\displaystyle CD}$ until they meet at the point of intersection, ${\displaystyle E}$. Let the midpoint of the segment ${\displaystyle EF}$ be ${\displaystyle N}$, and let the midpoint of the segment ${\displaystyle BC}$ be ${\displaystyle M}$ (Figure 1).

Theorem[]

If the midpoint of the line segment ${\displaystyle BF}$ is ${\displaystyle P}$, the Newton–Gauss line of the complete quadrilateral ${\displaystyle ABCDEF}$ and the line ${\displaystyle PM}$ determine an angle ${\displaystyle \angle PMN}$ equal to ${\displaystyle \angle EFD}$.

Proof[]

First show that the triangles ${\displaystyle \triangle NPM}$ and ${\displaystyle \triangle EDF}$ are similar.

Since ${\displaystyle BE\parallel PN}$ and ${\displaystyle FC\parallel PM}$, we know ${\displaystyle \angle NPM=\angle EAC}$. Also, ${\displaystyle \left({\frac {BE}{PN}}\right)}$${\displaystyle =\left({\frac {FC}{PM}}\right)=2.}$

In the cyclic quadrilateral ${\displaystyle ABCD}$, these equalities hold:

${\displaystyle \angle EDF=\angle ADF+\angle EDA=\angle ACB+\angle ABC=\angle EAC.}$

Therefore, ${\displaystyle \angle NPM=\angle EDF.}$

Let ${\displaystyle R_{1}}$ and ${\displaystyle R_{2}}$ be the radii of the circumcircles of ${\displaystyle \triangle EDB}$ and ${\displaystyle \triangle FCD}$, respectively. Apply the law of sines to the triangles, to obtain:

${\displaystyle {\frac {BE}{FC}}={\frac {2R_{1}\sin EDB}{2R_{2}\sin FDC}}={\frac {R_{1}}{R_{2}}}={\frac {2R_{1}\sin EBD}{2R_{2}\sin FCD}}={\frac {DE}{DF}}.}$

Since ${\displaystyle BE=2PN}$and ${\displaystyle FC=2PM}$, this shows the equality ${\displaystyle {\frac {PN}{PM}}={\frac {DE}{DF}}.}$ The similarity of triangles ${\displaystyle PMN}$ and ${\displaystyle DFE}$ follows, and ${\displaystyle \angle NMP=\angle EFD.}$

Remark[]

If ${\displaystyle Q}$ is the midpoint of the line segment ${\displaystyle FC}$, it follows by the same reasoning that ${\displaystyle \angle NMQ=\angle EFA.}$

Figure 2: Isogonal lines.

Isogonal lines[]

Theorem[]

The line through ${\displaystyle E}$ parallel to the Newton–Gauss line of the complete quadrilateral ${\displaystyle ABCDEF}$ and the line ${\displaystyle EF}$ are isogonal lines of ${\displaystyle \angle BEC}$, that is, each line is a reflection of the other about the angle bisector.^[5] (Figure 2)

Proof[]

Triangles ${\displaystyle EDF}$ and ${\displaystyle NPM}$ are similar by the above argument, so ${\displaystyle \angle DEF=\angle PNM}$. Let ${\displaystyle E'}$ be the point of intersection of ${\displaystyle BC}$ and the line parallel to the Newton–Gauss line ${\displaystyle NM}$ through ${\displaystyle E}$.

Since ${\displaystyle PN||BE}$ and ${\displaystyle NM||EE'}$, ${\displaystyle \angle BEF=\angle PNF}$, and ${\displaystyle \angle FNM=\angle E'EF}$.

Therefore, ${\displaystyle \angle CEE'=\angle DEF-\angle E'EF=\angle PNM-\angle FNM=\angle PNF=\angle BEF.}$

Two cyclic quadrilaterals sharing a Newton-Gauss line[]

Figure 3: Showing that the quadrilaterals MPGN and MQHN are cyclic.

Lemma[]

Let ${\displaystyle G}$ and ${\displaystyle H}$ be the orthogonal projections of the point ${\displaystyle F}$ on the lines ${\displaystyle AB}$ and ${\displaystyle CD}$ respectively.

The quadrilaterals ${\displaystyle MPGN}$ and ${\displaystyle MQHN}$are cyclic quadrilaterals.^[5]

Proof[]

${\displaystyle \angle EFD=\angle PMN}$, as previously shown. The points ${\displaystyle P}$ and ${\displaystyle N}$ are the respective circumcenters of the right triangles ${\displaystyle \triangle BFG}$ and ${\displaystyle \triangle EFG}$. Thus, ${\displaystyle \angle PGF=\angle PFG}$ and ${\displaystyle \angle FGN=\angle GFN}$.

Therefore,

${\displaystyle {\begin{aligned}\angle PGN+\angle PMN&=(\angle PGF+\angle FGN)+\angle PMN\\[4pt]&=\angle PFG+\angle GFN+\angle EFD\\[4pt]&=180^{\circ }\end{aligned}}.}$

Therefore, ${\displaystyle MPGN}$ is a cyclic quadrilateral, and by the same reasoning, ${\displaystyle MQHN}$ also lies on a circle.

Figure 4: Showing that the complete quadrilaterals EDGHIJ and ABCDEF have the same Newton–Gauss line.

Theorem[]

Extend the lines ${\displaystyle GF}$ and ${\displaystyle HF}$ to intersect ${\displaystyle EC}$ and ${\displaystyle EB}$ at ${\displaystyle I}$ and ${\displaystyle J}$ respectively (Figure 4).

The complete quadrilaterals ${\displaystyle EFGHIJ}$ and ${\displaystyle ABCDEF}$ have the same Newton–Gauss line.^[5]

Proof[]

The two complete quadrilaterals have a shared diagonal, ${\displaystyle EF}$. ${\displaystyle N}$ lies on the Newton–Gauss line of both quadrilaterals. ${\displaystyle N}$ is equidistant from ${\displaystyle G}$ and ${\displaystyle H}$, since it is the circumcenter of the cyclic quadrilateral ${\displaystyle EGFH}$.

If triangles ${\displaystyle \triangle GMP}$ and ${\displaystyle \triangle HMQ}$ are congruent, and it will follow that ${\displaystyle M}$ lies on the perpendicular bisector of the line ${\displaystyle HG}$. Therefore, the line ${\displaystyle MN}$ contains the midpoint of ${\displaystyle GH}$, and is the Newton–Gauss line of ${\displaystyle EFGHIJ}$.

To show that the triangles ${\displaystyle \triangle GMP}$ and ${\displaystyle \triangle HMQ}$ are congruent, first observe that ${\displaystyle PMQF}$ is a parallelogram, since the points ${\displaystyle M}$ and ${\displaystyle P}$ are midpoints of ${\displaystyle BF}$ and ${\displaystyle BC}$ respectively.

Therefore,

• ${\displaystyle MP=QF=HQ,}$
• ${\displaystyle GP=PF=MQ,}$ and
• ${\displaystyle \angle MPF=\angle FQM.}$

Also note that

${\displaystyle \angle FPG=2\angle PBG=2\angle DBA=2\angle DCA=2\angle HCF=\angle HQF.}$

Hence,

${\displaystyle \angle MPG=\angle MPF+\angle FPG=\angle FQM+\angle HQF=\angle HQF+\angle FQM=\angle HQM.}$

Therefore, ${\displaystyle \triangle GMP}$ and ${\displaystyle \triangle HMQ}$ are congruent by SAS.

Remark[]

Due to ${\displaystyle \triangle GMP}$ and ${\displaystyle \triangle HMQ}$ being congruent triangles, their circumcircles ${\displaystyle MPGN}$ and ${\displaystyle MQHN}$ are also congruent.

History[]

The Newton–Gauss line proof was developed by the two mathematicians it is named after: Sir Isaac Newton and Carl Friedrich Gauss.^{[citation needed]} The initial framework for this theorem is from the work of Newton, in his previous theorem on the Newton line, in which Newton showed that the center of a conic inscribed in a quadrilateral lies on the Newton–Gauss line.^[6]

The theorem of Gauss and Bodenmiller states that the three circles whose diameters are the diagonals of a complete quadrilateral are coaxal.^[7]

Notes[]

References[]

• Johnson, Roger A. (2007) [1929], Advanced Euclidean Geometry, Dover, ISBN 978-0-486-46237-0
  • (available on-line as) Johnson, Roger A. (1929). "Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle". HathiTrust. Retrieved 28 May 2019.

External links[]

• Bogomonly, Alexander. "Theorem of Complete Quadrilateral: What is it?". Retrieved 11 May 2019.