Pappus's Hexagon Theorem

Pappus's (Hexagonal) Theorem

Now we will look at Pappus's theorem. It states that

Suppose the vertices of a hexagon alternately lie on two lines. Then the three pairs of opposite sides intersect at points that are collinear.

This is a specific case of Pascal's Theorem.

Proof of the Theorem

The proof that I am going to present will require some knowledge in projective geometry.

Lemma 1: A projectivity that maps a range to another is a perspectivity if and only if the intersection of the two lines are invariant under projectivity.

Proof: The intersection is obviously invariant under projectivity. Suppose that a projectivity leaves the intersection invariant. Let $A$ and $B$ be two points in the first range and $A'$ and $B'$ be the two corresponding points. Let $O$ be the intersection of $AA'$ and $BB'$. Then by the fundamental theorem of projective geometry, we have that the perspectivity mapping $ABE$ to $A'B'E$ is the same as the projectivity we defined. $\Box$

Now time for the proof of the theorem.

Proof of the Theorem: Suppose we have hexagon $A_1B_2C_1A_2B_1C_2$, as shown below. Let $H$ be the intersection of $A_1B_2$ and $A_2C_1$, $I$ be the intersection of $A_1C_2$ and $B_2C_1$, and $G$ be the intersection of $A_1B_1$ and $A_2B_2$.

Notice that $A_1FHB_2 \doublebarwedge A_1B_1C_1G$ from point $A_2$ and $A_1B_1C_1G \doublebarwedge IDC_1B_2$ from point $C_2$ so there is a unique projectivity from $A_1FH$ and $IDC_1$. Since $B_2$, the intersection of the two lines, is invariant under the projectivity, we have that it is a perspectivity from $E$. Thus $D$, $E$, and $F$ are collinear. $\Box$ References

1. H. S. M. Coxeter, Projective Geometry, Springer-Verlag, 1987