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Can you provide a proof for the following claim:
Claim. Construct a convex heptagon circumscribed about an ellipse. Intersection points of its non-adjacent longer diagonals lie on a common ellipse.
GeoGebra applet that demonstrates this claim can be found here.
I have noticed that principal diagonals of the hexagon defined by the points: $𝐺,FG \cap ED,𝐷,𝐶,𝐵,𝐴$ concur at the point $P$. Similar is true for the points $Q,R,S,M,N,O$ as well . So I guess we should apply Brianchon's theorem somehow.
The comments mention the Heptagon Theorem:
Let $H$ be a heptagon with seven vertices given in cyclic order inscribed in a conic. Then the Pascal lines of the seven hexagons obtained by omitting each vertex of $H$ in turn and keeping the remaining vertices in the same cyclic order are the sides of a heptagon $I$ which circumscribes a conic.
Moreover, the Brianchon points of the seven hexagons obtained by omitting the sides of $I$ one at a time and keeping the remaining sides in the natural cyclic order are the vertices of the original heptagon.
Since this is a theorem about points, lines, conics, and incidence, it is a theorem of projective geometry and can be dualized. The dual of the first paragraph is
Let $H$ be a heptagon with seven sides given in cyclic order which circumscribes a conic. Then the Brianchon points of the seven hexagons obtained by omitting each side of $H$ in turn and keeping the remaining sides in the same cyclic order are the vertices of a heptagon $I$ inscribed in a conic.
As noted in the OP, the Brianchon points of the seven hexagons are the pairwise intersections of consecutive long diagonals (by which we mean their endpoints are in cyclic order, such as $AD,BE$). So the dualized first paragraph proves the OP's claim.
The Heptagon theorem is stated and proven in Evelyn et al, The Seven Circles Theorem and Other New Theorems (if you can't get access to the book, the proof is shown below). As done above, the Heptagon theorem can be dualized, or the proof of the Heptagon Theorem can be dualized, which is a lot more work.
It's worth pointing out, if it isn't already obvious, that the Heptagon Theorem contains dual constructions. Starting with an inscribed heptagon one uses Pascal lines to construct a circumscribed heptagon. One can then reverse the construction using Brianchon points to construct an inscribed heptagon.
The picture given in the Mathworld article uses a doubly winding blue heptagon to obtain a smaller red heptagon. But, if you start with a singly winding heptagon, you end up with a red heptagon typically much larger than the blue heptagon. For example the Pascal lines of the orange heptagon $MNOPQRS$ in the OP are the sidelines of the blue heptagon $ABCDEFG$.
Below is a copy of the proof of the Hexagon Theorem.
