Geometria Super Top, Selçuk Biçer, https://www.facebook.com/photo/?fbid=28744356998487035&set=p.28744356998487035&opaqueCursor=AbogKnAhG3y26cGgbz3XCpxwBMVeB940XvB2Wob-n2IojCvzqxs3E-OR5myu241jnOmbOADLsIiYW_NKSgxW9akF-jLASW1LM5ghSZ_EteggbEUCejMdpAzqpTpOOb45e29E-Fr9BieuWgPj6pYMZz7L8rQBiEETczHqSaf3G19lOmtZzI8uzz6VqokW_fGWNsw90ZBh1hxszFqvRrnhI-_EFEy4RZZFKFHmm8fk1Bqy-ansUTV9OY80h3NnnjUfpKJ86TTIWN4igjpKgD5AKC78WCmiG4pMqQzGjv8Q5fLObU-6287yKKDAA4FDM0D7O7jHJ3vRWS7LT-a5jlYYiZckZkiVMBNA-prxpHrvP0Vq_J6cugV4997HfD0rbTz07DXNDSpHFhtIURKOkE6cxEbeqzmVHKrPUOe2gVwCvP-dpwm4llliXQ2QUNBfZQXCVkZa_ZSoUcRDtgCRmR1Zf9pBcnQ50lsDDqjOAifcZo3w1wEUt9lJAYJEzJzS5PttWydDIGhk2XMzBGb6tc1WY7Pd2KSbhjbwCzweLSMXQVf7VCoUljFyN5enBl8NBSI014k

 

Problem posted by Wellyngton Silva on the page 'Geometria Super Top'

 

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.

https://gogeometry.blogspot.com/2026/07/geometric-challenge-1621-problem.html

 

Mathstackexchange, geometry - In $\triangle ABC$, $AB = AC$ and $\angle B = 40^\circ$ . $D$ is a point on $AB$ produced such that $AD = BC$. Join $DC$. Find $\angle DCB$ . - Mathematics Stack Exchange

 

Angel Silva Palacios, Training Geometry Olympiad, https://www.facebook.com/photo?fbid=1872372997056216&set=gm.1694422275151412&idorvanity=293155508611436

 

Anie Luc, Geometria Super Top, https://www.facebook.com/photo?fbid=122159419298416607&set=gm.2323356811768008&idorvanity=830782544358783

 

Geometria Super Top, Chris Wildhagen, problem by Anie Luc, https://www.facebook.com/photo/?fbid=1791241638703011&set=gm.2322429151860774&idorvanity=830782544358783

 

Training Geometry Olympiad, Angel Silva Palacios, https://www.facebook.com/photo?fbid=1867830004177182&set=gm.1690140702246236&idorvanity=293155508611436

 

Training Geometry Olympiad, Angel Silva Palacios, https://www.facebook.com/photo?fbid=1863384657955050&set=gm.1685936825999957&idorvanity=293155508611436

 

Thiago Durado, ‘I Love Geometry’ https://www.facebook.com/photo?fbid=1284217157212573&set=gm.1722819092488731&idorvanity=1216680106435968

 

https://artofproblemsolving.com/community/c4t48f4h2412508_midpoint_of_xy_lies_on_diagonal_bd_of_square_abcd_2020_nzmo_12

 

Abdilkadir Altıntaş, ‘Geometri Gűnlűǧű’, Problem 159

  A problem from Abdilkadir Altıntaş, ‘Geometri Gűnlűǧű’, Problem 159:   

Gogeometry https://gogeometry.blogspot.com/2008/05/geometry-problem-2.html

 

Go-geometry, https://gogeometry.blogspot.com/2008/05/geometry-problem-1.html

 

Artofproblemsolving.com https://artofproblemsolving.com/community/c4t48f4h3864066_very_very_nice

 

Artofproblemsolving https://artofproblemsolving.com/community/c4t48f4h3860464_bridging_the_gap_geo_problem

 

Trazo Maldito, MatheBox, https://www.facebook.com/photo/?fbid=1460236849231293&set=pcb.1460236909231287