Proof for Brianchon's theorem I seek a proof of Brianchon's theorem, which states:

When a hexagon is circumscribed around a conic section, its principal diagonals (those connecting opposite vertices) meet in a single point.

I'd prefer simpler proofs than more complex ones. Thank you.
(P.S: Similar questions have been asked, but none received proper answers.)
 A: I worked out a proof using Ceva's theorem, but it requires some obscure supplementary lemmas that make me doubt that the proof would be as simple as OP is hoping.  Quoting Cut The Knot:

The easiest way to prove Brianchon's theorem is by way of duality
implied by the properties of poles and polars.

But that is just a way of reducing it to Pascal's theorem, which is fine if you're happy with not having to prove Brianchon's theorem directly.
An elegant modern proof (Coxeter attributes it to Smogorzhevskiĭ) is given on Pamfilos' site.
My version of a Ceva proof was guided by exercises 159-161 in Smith's Modern Geometry.  But there are several proofs in Hatton's Projective Geometry, Chapter XIV.  They are laid out side by side with proofs of Pascal's Theorem, including a Ceva proof. (Proof by Carnot's Theorem, pg 191.  Aptly named because a dual of Carnot's Theorem, attributed to Chasles, is the crux of the proof).  These proofs are fully general for conics, but the proofs for the circle case are presented in Chapter XII.
Milne (Cross-Ratio Geometry, pg 158) gives a nice short proof, but as usual it requires some understanding of projective geometry.
