Theorems of Pappus and Pascal

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 \(C_2=C_2^{\prime} \).

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 \(C_2=C_2^{\prime} \).