若$A, B, C', B'$ not in general position.(其中有三点共线):
设 $C'∈AB$, 则$B C'∩B' C=B C'∩B' C=C$, 此时$A B'∩A' B,C,C$共线.
下面设$A, B, C', B'$ in general position(其中任意三点不共线):
根据Theorem 3, 建立射影坐标$[x_0,x_1,x_2]$\begin{aligned}
A & =[1,0,0] \\
B & =[0,1,0] \\
C' & =[0,0,1] \\
B' & =[1,1,1]
\end{aligned}The line $A B$ is defined by the 2-dimensional subspace $\left\{\left(x_0, x_1, x_2\right) \in F^3: x_2=0\right\}$, so the point $C$, which lies on this line, is of the form $C=[1, c, 0]$ and $c \neq 0$ since $A \neq C$. Similarly the line $B' C'$ is $x_0=x_1$, so $A'=[1,1, a]$ with $a \neq 1$.
The line $B C'$ is defined by $x_0=0$ and $B'C$ is defined by the span of $(1,1,1)$ and $(1, c, 0)$, so the point $B C'∩B'C$ is represented by the linear combination of $(1,1,1)$ and $(1, c, 0)$ for which $x_0=0$, i.e.
$$
(1,1,1)-(1, c, 0)=(0,1-c, 1)
$$
The line $C'A$ is given by $x_1=0$, so similarly $C A'∩C'A$ is represented by
$$
(1, c, 0)-c(1,1, a)=(1-c, 0,-c a)
$$
Finally $A B'$ is given by $x_1=x_2$, so $A B'∩A'B$ is
$$
(1,1, a)+(a-1)(0,1,0)=(1, a, a)
$$
But then
$$
(c-1)(1, a, a)+(1-c, 0,-c a)+a(0,1-c, 1)=0 .
$$
Thus the three vectors span a 2-dimensional subspace and so the three points lie on a projective line.
