# Theorems of Pappus and Pascal

** Theorem 4 (Pappus) **

The points \( A_1 \), \( A_2 \), \( A_3 \) belong to the line \( a \),
and the points \( B_1 \), \( B_2 \), \( B_3 \) belong to the line \( b \).
Assume that
\( A_1B_2\cap A_2B_1=C_3 \), \( A_1B_3\cap A_3B_1=C_2 \),
\( A_2B_3\cap A_3B_2=C_1 \). Then \( C_1 \), \( C_2 \), \( C_3 \)
are colinear.

Denote \( C_2^{\prime} =C_1C_3\cap A_3B_1 \), \( D=A_1B_2\cap A_3B_1 \),
\( E=A_2B_1\cap A_3B_2 \), \( F=a\cap b \).
Our goal is to prove that the points \( C_2 \) and
\( C_2^{\prime} \) are identical.

Consider the sequence of projectivities:
\[ A_3B_1DC_2\frac{A_1}{\overline\wedge}FB_1B_2B_3
\frac{A_2}{\overline\wedge}A_3EB_2C_1\frac{C_3}{\overline\wedge}A_3B_1DC_2^{\prime} .\]
We have got the projective transformation of the line
\( A_3B_1 \) that fixes the points \( A_3 \), \( B_1 \), \( D \), and maps
\( C_2 \) to \( C_2^{\prime} \). Since the projective mapping with
three fixed points is the identity we have
\( C_2=C_2^{\prime} \).

** Theorem 5 (Pascal) **

Assume that the points
\( A_1 \), \( A_2 \), \( A_3 \), \( B_1 \), \( B_2 \), \( B_3 \) belong to a
circle. The point in intersections of \( A_1B_2 \) with \( A_2B_1 \),
\( A_1B_3 \) with \( A_3B_1 \), \( A_2B_3 \) with \( A_3B_2 \)
lie on a line.

The points \( C_2^{\prime} \), \( D \), and \( E \)
as in the proof of the Pappus theorem.

Consider the sequence of perspectivities
\[ A_3B_1DC_2\frac{A_1}{\overline\wedge}A_3B_1B_2B_3\frac{A_2}{\overline\wedge}A_3EB_2C_1
\frac{C_3}{\overline\wedge}A_3B_1DC_2^{\prime} .\] In the same way as above we conclude
that \( C_2=C_2^{\prime} \).