Pascal's (Hexagon) Theorem

Pascal's Hexagon Theorem says that if you have a hexagon $ABCDEF$ that lies on a conic, the intersections of the three pairs of opposite sides are collinear. Formally, if $X=AB\cap DE, Y=BC\cap EF$ and $Z=CD\cap FA$, then $X,Y,Z$ are collinear.

The proof of this fact follows quickly using cross ratios. Let $P = BC\cap DE, Q = AB\cap CD, Y' = XZ\cap BC$. It is sufficient to show that $Y = Y'$

Note that $$(C, Y; P, B) \overset{E}{=} (C, F; D, B) \overset{A}{=} (C, Z; Q, D) \overset{X}{=} (C, Y'; P, B)$$

so $Y = Y'$ as desired.

Note that in this theorem, the hexagon $ABCDEF$ does not have to be convex. In fact, it does not even have to be a hexagon! Consider this hexagon...

If we draw the intersections...

The theorem is indeed still true. $X,Y,Z$ are collinear!

Furthermore, the points do not even have to be distinct! Pascal's still gives nontrivial results if one, two, or even three points are repeated. For example, with points $AABBCC$, we would have that $AA \cap BC, AB \cap CC, BB \cap CA$ are collinear, where $AA, BB, CC$ represent the tangents to the circumcircle of $\triangle ABC$ at points $A, B, C$ respectively.

This particular case gives the following result, called the Lemoine Line, which has an otherwise nontrivial proof:

Note to writers: Someone please add the intersections between the tangents from $A$ and $BC$, $B$ and $CA$, $C$ and $AB$, and draw a line between these points.