# proof of Pappus’s theorem

Pappus’s theorem says that if the six vertices of a hexagon lie alternately on two lines, then the three points of intersection of opposite sides are collinear. In the figure, the given lines are $A_{11}A_{13}$ and $A_{31}A_{33}$, but we have omitted the letter $A$.

The appearance of the diagram will depend on the order in which the given points appear on the two lines; two possibilities are shown.

Pappus’s theorem is true in the affine plane over any (commutative) field. A tidy proof is available with the aid of homogeneous coordinates.

No three of the four points $A_{11}$, $A_{21}$, $A_{31}$, and $A_{13}$ are collinear, and therefore we can choose homogeneous coordinates such that

 $A_{11}=(1,0,0)\qquad A_{21}=(0,1,0)$
 $A_{31}=(0,0,1)\qquad A_{13}=(1,1,1)$

That gives us equations for three of the lines in the figure:

 $A_{13}A_{11}:y=z\qquad A_{13}A_{21}:z=x\qquad A_{13}A_{31}:x=y\;.$

These lines contain $A_{12}$, $A_{32}$, and $A_{22}$ respectively, so

 $A_{12}=(p,1,1)\qquad A_{32}=(1,q,1)\qquad A_{22}=(1,1,r)$

for some scalars $p,q,r$. So, we get equations for six more lines:

 $A_{31}A_{32}:y=qx\qquad A_{11}A_{22}:z=ry\qquad A_{12}A_{21}:x=pz$ (1)
 $A_{31}A_{12}:x=py\qquad A_{11}A_{32}:y=qz\qquad A_{21}A_{22}:z=rx$ (2)

By hypothesis, the three lines (1) are concurrent, and therefore $prq=1$. But that implies $pqr=1$, and therefore the three lines (2) are concurrent, QED.

Title proof of Pappus’s theorem ProofOfPappussTheorem 2013-03-22 13:47:40 2013-03-22 13:47:40 mathcam (2727) mathcam (2727) 6 mathcam (2727) Proof msc 51A05