proof of Pappus’s theorem

Pappus’s theorem says that if the six vertices of a hexagon liealternately on two lines, thenthe 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 thegiven 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 $A_{13}$ arecollinear, 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.