The Theorem of Pappus

Interactive Course on Projective Geometry

In the geometric constructions that follow, we will denote points by uppercase letters and lines by lowercase letters. The symbol $ AB$ denotes the line passing through the points $ A$ and $ B$, while $ a \cdot b$ denotes the point of intersection of the lines $ a$ and $ b$.

Theorem 0.1 (Pappus)   Draw two lines on the projective plane and three points on each line. Denote the points on the first line by $ A,B, C$, and the points on the second line by $ A',B',C'$. Draw the lines that join points denoted by different letters (i.e., we do not draw the lines $ AA', BB'$, or $ CC'$). The points $ AB'\cdot BA', AC' \cdot CA'$, and $ BC' \cdot CB'$ are collinear.

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The proof is given in the following two exercises, the first of which is to prove the affine version of the theorem.

Exercise 0.1 (10)   Draw two lines on the plane and draw points $ A,B, C$ on the first line, and points $ A',B',C'$ on the second line such that $ AB'$ is parallel to $ BA'$, and $ BC'$ is parallel to $ CB'$. Show that $ AC'$ is necessarily parallel to $ CA'$.

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Now comes the reduction of the projective result to the affine result.

Exercise 0.2 (10)   Let $ {\cal T}$ be any projective transformation that sends the points $ AB'\cdot BA'$ and $ BC' \cdot CB'$ to infinity. Apply $ {\cal T}$ to the points and lines in the configuration of theorem 0.1 and verify that the new configuration is like the one of the previous exercise. Use this to prove Pappus' theorem.