Newton–Gauss line

Labels used in proof concerning complete quadrilateral

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', 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,

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

However, the line △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, N are collinear on the sides of triangle △PQR.

Figure 1: An angle equality.

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).

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.

First show that the triangles △NPM, △EDF are similar.

Since BE ∥ PN and FC ∥ PM, we know ∠NPM = ∠EAC. Also, ${\displaystyle {\tfrac {\overline {BE}}{\overline {PN}}}={\tfrac {\overline {FC}}{\overline {PM}}}=2.}$

In the cyclic quadrilateral ABCD, these equalities hold:

${\displaystyle {\begin{aligned}\angle EDF&=\angle ADF+\angle EDA,\\&=\angle ACB+\angle ABC,\\&=\angle EAC.\end{aligned}}}$

Therefore, ∠NPM = ∠EDF.

Let R₁, R₂ be the radii of the circumcircles of △EDB, △FCD respectively. Apply the law of sines to the triangles, to obtain:

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

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

If Q is the midpoint of the line segment FC, it follows by the same reasoning that ∠NMQ = ∠EFA.

Figure 2: Isogonal lines.

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

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

Since PN ∥ BE and NM ∥ EE', ∠BEF = ∠PNF, and ∠FNM = ∠E'EF.

Therefore,

${\displaystyle {\begin{aligned}\angle CEE'&=\angle DEF-\angle E'EF,\\&=\angle PNM-\angle FNM,\\&=\angle PNF=\angle BEF.\end{aligned}}}$

Let G and H be the orthogonal projections of the point F on the lines AB and CD respectively.

The quadrilaterals MPGN and MQHN are cyclic quadrilaterals.[5]

∠EFD = ∠PMN, as previously shown. The points P and N are the respective circumcenters of the right triangles △BFG, △EFG. Thus, ∠PGF = ∠PFG and ∠FGN = ∠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, MPGN is a cyclic quadrilateral, and by the same reasoning, MQHN also lies on a circle.

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

Extend the lines GF, HF to intersect EC, EB at I, J respectively (Figure 4).

The complete quadrilaterals EFGHIJ and ABCDEF have the same Newton–Gauss line.[5]

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

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

To show that the triangles △GMP, △HMQ are congruent, first observe that PMQF is a parallelogram, since the points M, P are midpoints of BF, BC respectively.

Therefore,

${\displaystyle {\begin{aligned}&{\overline {MP}}={\overline {QF}}={\overline {HQ}},\\&{\overline {GP}}={\overline {PF}}={\overline {MQ}},\\&\angle MPF=\angle FQM.\end{aligned}}}$

Also note that

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

Hence,

${\displaystyle {\begin{aligned}\angle MPG&=\angle MPF+\angle FPG,\\&=\angle FQM+\angle HQF,\\&=\angle HQF+\angle FQM,\\&=\angle HQM.\end{aligned}}}$

Therefore, △GMP and △HMQ are congruent by SAS.

Due to △GMP, △HMQ being congruent triangles, their circumcircles MPGN, MQHN are also congruent.