For a degenerate conic that is two intersecting lines, the tangents are the pencil of lines going through the intersection. So the only Brianchon hexagon will have vertices that are identical.

The diagram you present can be thought of as a Brianchon hexagon for the degenerate conic that is the two points $G$ and $H$. It is a *line conic*, i.e. a conic that is defined as the envelope of a set of tangents. The tangents in this case are the two pencils of lines going through $G$ and $H$. And $G$ and $H$ can be thought of as the points that are dual to the outer straight lines in Pappus' Theorem. The mapping of a point conic to a line conic is an essential part of the dualization of Pappus' Theorem.

The figure above is from Richter-Gilbert, *Perspectives on Projective Geometry*, pg 161 and shows a progression of line conics going from ellipse to hyperbola, going through the two point line conic.

A good place to start is this blog post on Dandelin's proof of Brianchon's Theorem.

A single sheeted hyperboloid is a ruled surface containing two sets of lines. The blog post calls them A-lines and B-lines, but we may as well call them red and blue lines. Any given pair of red and blue lines will either meet or be parallel to one another.

Pick six lines, three of each color. They will meet up to form a non-planar hexagon, called a skew hexagon.

The diagram below, taken from the blog post, shows such a skew hexagon with alternating red and blue edges and the three yellow diagonals. The diagonals are shown to be concurrent, mainly because three planes intersect in a point.

In sections 13-15 of the translation Dandelin defines an angle plane as the plane formed at a vertex by its incident edges. He shows that, for a skew hexagon, angle planes of opposite vertices meet in a line, and that these three lines lie in a common plane $p$.

In section 16, Dandelin discusses the conic formed by intersecting the hyperboloid with a plane $q$. He selects six points on this conic, which form a planar hexagon. Each point has a red and blue line running through it, and Dandlelin chooses alternating red and blue lines (called direct and inverse). These lines form a skew hexagon. The vertices of the skew hexagon correspond to the sides of the planar hexagon. Indeed, each side of the planar hexagon lies on the angle plane of the corresponding skew hexagon vertex.

Finally, he considers opposite sides of the planar hexagon. They will meet on the intersection lines of the corresponding angle planes. But because these lines are on a common plane $p$, the opposite sides will meet on the line that is the intersection of the common plane $p$ and the conic plane $q$.

Brilliant.

He won a prize for this (from Ruff, *An Extension of Pascal's Theorem*)

In 1825 the Académie Royale de Bruxelles proposed as a prize topic the
extension of Pascal's theorem to space of three dimensions. The prize
was won by Dandelin, who showed that a skew hexagon formed of three
lines from each regulus of a hyperboloid of revolution has the Pascal
property that pairs of opposite planes meet on a plane, and the dual,
or Brianchon property, that the lines joining pairs of opposite
vertices meet in a point.

Hopefully, pondering this answer, along with the blog post and the Dandelin translation, will clarify what's going on. (The original 1826 paper is here.)

## Best Answer

Your proof is already pretty short, but it can be shorter. For the hexagon $\pmb{H}=AGBDJE$ you've shown that the diagonals $AD,GJ,BE$ are concurrent. By the converse of Brianchon, $\pmb{H}$ has an inscribed conic. But the sides of $\pmb{H}$ are the same as the sides of the triangles, so we're done.