Projective Collinearity in a Quadrilateral: What Is This About? A Mathematical Droodle

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Points K, L, M, N are selected arbitrarily on the sides AB, BC, CD, and AD of a quadrilateral ABCD. Let P denote the intersection of the diagonals AC and BD and extend the lines PK, PL, PM, PN beyond P to their intersection with the opposite sides. Denote the points of intersection K', L', M', and N', so that the lines KK', LL', MM', and NN' all meet in P. Finally, let E be the intersection of KM and LN and E' be the intersection of K'M' and L'N'. Then the three points E, P, and E' are collinear.

Proof

Introduce two new points: X the intersection of a pair of opposite sides AB and CD and Y the intersection of the other pair of opposite sides BC and AD. If the side lines in either pair are parallel then the corresponding point lies at infinity. If both X and Y lie at infinity then AB||CD and BC||AD, so that the quadrilateral ABCD is parallelogram and P is its center of symmetry: K' is symmetric with K, L' with L, and so on. As an important consequence, E and E' are also symmetric in P. The assertion that E, P, and E' is quite obvious in this case.

If either X or Y (or both) is finite, a projective transformation could be used to move the line XY to infinity. Then the image of ABCD will become a parallelogram, the images of E, P, and E' will be found to be collinear, which might only be possible if the points E, P, E' were collinear in the first place.

References

1. A. A. Zaslavsky, The Orthodiagonal Mapping of Quadrilaterals, Kvant, n 4, 1998, pp 43-44 (in Russian), pdf is available at https://kvant.mccme.ru/1998/04/. • 