## Symmetric hexagons

This is a subject in which participate three basic theorems of projective geometry. The theorems of Pascal, Brianchon and Desargues (French trio). Consider a conic (c) and four points A, B, C, D on it. Then define E by intersecting (c) with the parallel to AB from D and finally point F by intersecting (c) with the parallel to BC from E. By Pascal's theorem FA will be parallel to CD, since the so created hexagon ABCDEF is inscribed in (c) and two of the three pairs of opposite sides intersect on the line at infinity.
Draw now the conic (c') tangent to the first five sides of the hexagon: {AB, BC, CD, DE, EF}. Conic (c') is also tangent to the sixth side FA exactly when the lines connecting opposite vertices meet at a point. This is again a consequence of Brianchon's theorem.
Now, if the opposite sides are parallel and lines joining opposite vertices pass through a point O, then O is center of symmetry of the hexagon. To see this apply Desarque's theorem on two opposite triangles, like for example BCD and EFA. The triangles being point-perspective (lines joining corresponding vertices pass through the same point = perspectivity center), they are also line-perspective (intersections of corresponding opposite sides are on a line). But the two pairs of opposite sides {BC,EF} and {CD,FA} intersect at the line at infinity. Hence the third pair {BD, EA} intersects also there, i.e. the lines are parallel. Thus, ABDE is a parallelogram and its diagonals are bisected at their intersection.

Disregarding the intervening conics we arive at the result, that a hexagon with parallel sides is symmetric exactly when the lines joining opposite vertices pass through a point, which is then the center of symmetry.

The file HexagonsSymmetricRepeat.html displays a figure and some questions concerning an infinite sequence of hexagons, one inscribed into the other, initiated by an arbitrary symmetric hexagon.