Few relations within a quadrilateral

 

 “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.