Note: I came up with the following proof 53 years ago, during an exam; as the majority of solvers used the 3 concurrent circumcircles of the equilateral triangles and their centers, I doubt my proof was correctly appreciated, because I only got 15 pts out of 20, while the math work was perfect, and the physics one of at least 7 of 10!
Use 'double angle lemma', which you may find on this blog.
“Let ABCD be a convex quadrilateral, {E} =
AB∩CD, {F} = BC∩AD.
Then the
following equality occurs: 
”
To achieve
this goal we denote M, N and P the midpoints of the segments |AC|, |BD| and
|EF| respectively and shall provide the following proof to the fact that the
above three points are collinear ( Newton – Gauss straight line ):
Draw
parallel lines to CE through A and F and to AF through E and C and we get few
parallelograms. Lets denote X , Y, Z and T the reflections of D across M, P,
across the middle of the segment |AE|, and across the middle of the segment
|CF| respectively.
Denominate then
{B’}=AE∩XY and apply Menelaus theorem in the triangle XYZ with the transversal
EAB’, getting: 
. But from the parallelograms we get AZ=DE, AX=DC, EY=DF,
ZE=AD, hence 
( 1 ). If we denote {B”}=XY∩CF working the same
way in the triangle XTY with the transversal FCB” we get 
( 2 ), from the equalities (1) and (2) resulting
that AE and CF intersect XY at the same point which is, obviously, B, therefore
( 3 ). As B, X and Y
are the reflections of D across N, M and P respectively, it follows that M, N
and P are collinear as well and 
, hence 
( 4 ).
Following
the same procedure around B with parallel lines through A and E to BF and
through C and F to BE we shall get 
( 5 ).
Applying a
well known property of equal ratios, from equalities (4) and (5) we get 
( 6 ) or 
( 6’ ).
Working
then around points A and C we shall get 
( 7 ).
Note that
in (6’) and (7) the left ratios have same denominator and the difference of
nominators equals the denominator, hence 
, what we intended to prove. The value of the difference is
positive or negative, depending upon the sequence of M, N and P on the straight
line.
Note: I came up with the following proof 53 years ago, during an exam; as the majority of solvers used the 3 concurrent circumcircles of th...
