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