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  "body": "\n## Statement\n\nLet $k$ be a [[field|04C5ZBG1WT]], and work in the [[projective plane|DANFNN8C4Y]] $\\mathbb{P}^2(k)$.\nLet $g$ and $h$ be two distinct lines meeting at a point $O$.\nLet $A, B, C$ be three distinct points of $g$ and $a, b, c$ three distinct points of $h$, all different from $O$.\nThen the three points\n$$X = Ab \\cap aB, \\quad Y = Ac \\cap aC, \\quad Z = Bc \\cap bC$$\nare [[collinear|0AFNNTRTJE]].\n\n## Remarks\n\nHere $Ab$ denotes the line through $A$ and $b$; the two lines in each intersection are distinct, so $X$, $Y$, $Z$ are well defined.\nOne proof chooses coordinates in which $g$ and $h$ are coordinate lines and $O$ is a coordinate point.\nIt then computes $X$, $Y$, $Z$ and checks that the $3 \\times 3$ determinant of their coordinates vanishes, using $xy = yx$ for scalars.\nAnother proof views $g \\cup h$ as a degenerate conic and applies [[Pascal's theorem|94AKWRHND0]] in the limiting case.\n\nThe commutativity of $k$ is necessary: over a noncommutative division ring $D$, Pappus's theorem fails in $\\mathbb{P}^2(D)$ for suitable points.\nIn fact a projective plane satisfies Pappus's theorem exactly when it is isomorphic to $\\mathbb{P}^2(k)$ for a field $k$.\n\nExample: in the real affine plane let $g$ be $y = 0$ and $h$ be $y = 1$, which meet at infinity.\nTake $A = (0,0)$, $B = (1,0)$, $C = (2,0)$ and $a = (0,1)$, $b = (1,1)$, $c = (2,1)$.\nThen $X = (1/2, 1/2)$, $Y = (1, 1/2)$, $Z = (3/2, 1/2)$, and all three lie on $y = 1/2$.\n\n## Explanations\n\n### L1 human\n\nPut three points on one line and three points on another line.\nConnect them crosswise in pairs, like the laces of a shoe, and mark where the crossing lines meet.\nThe three marks always lie on a straight line.\nThe proof is a small piece of algebra, and the key step is that multiplication of numbers is commutative.\n\n### L2 human\n\nPappus's theorem is the configuration theorem that encodes commutativity of the coordinates.\nA 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|5Z19Y02MME]].\nFor finite planes the two notions agree, because a finite division ring is commutative by Wedderburn's little theorem.\nIt is the special case of [[Pascal's theorem|94AKWRHND0]] in which the conic degenerates into the pair of lines $g$ and $h$.\n",
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