proof of Pappus's theorem
ProofOfPappussTheorem
2003-07-26 00:11:30
2003-08-10 01:45:00
Proof
Pappus's theorem
0
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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 $A$.
\includegraphics{pappus}
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:
\begin{equation} \label{I1}
A_{31}A_{32}:y=qx\qquad A_{11}A_{22}:z=ry\qquad A_{12}A_{21}:x=pz
\end{equation}
\begin{equation} \label{I2}
A_{31}A_{12}:x=py\qquad A_{11}A_{32}:y=qz\qquad A_{21}A_{22}:z=rx
\end{equation}
By hypothesis, the three lines (\ref{I1}) are concurrent, and therefore
$prq=1$. But that implies $pqr=1$, and therefore the three lines
(\ref{I2}) are concurrent, QED.