Pappus's hexagon theorem

Remarks

Here $Ab$ denotes the line through $A$ and $b$; the two lines in each intersection are distinct, so $X$, $Y$, $Z$ are well defined. One proof chooses coordinates in which $g$ and $h$ are coordinate lines and $O$ is a coordinate point. It then computes $X$, $Y$, $Z$ and checks that the $3 \times 3$ determinant of their coordinates vanishes, using $xy = yx$ for scalars. Another proof views $g \cup h$ as a degenerate conic and applies Pascal's theorem in the limiting case.

The commutativity of $k$ is necessary: over a noncommutative division ring $D$, Pappus's theorem fails in $\mathbb{P}^2(D)$ for suitable points. In fact a projective plane satisfies Pappus's theorem exactly when it is isomorphic to $\mathbb{P}^2(k)$ for a field $k$.

Example: in the real affine plane let $g$ be $y = 0$ and $h$ be $y = 1$, which meet at infinity. Take $A = (0,0)$, $B = (1,0)$, $C = (2,0)$ and $a = (0,1)$, $b = (1,1)$, $c = (2,1)$. Then $X = (1/2, 1/2)$, $Y = (1, 1/2)$, $Z = (3/2, 1/2)$, and all three lie on $y = 1/2$.

Going deeper

For a student

Pappus's theorem is the configuration theorem that encodes commutativity of the coordinates. A plane in which it holds is called Pappian, and by Hessenberg's theorem every Pappian plane is Desarguesian, so it also satisfies Desargues's theorem. For finite planes the two notions agree, because a finite division ring is commutative by Wedderburn's little theorem. It is the special case of Pascal's theorem in which the conic degenerates into the pair of lines $g$ and $h$.