Proof

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.

Remark

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

Proof

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,

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

Proof

∠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.

Proof

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.

Remark

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