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利用Desargues构型理解交比在射影下不变

如图1所示,$C_1A_1A_2 \frac{C_3}{\wedge }C_2A_3A_{2}$,$C_1B_1B_2\frac{C_3}{\wedge} C_2B_3B_2$.则Desargues定理表明,点$A_2,B_2,Z$共线,其中$Z$是直线$A_1B_1$与直线$A_3B_3$的交点.因此
图1
由Menelaus定理,
\begin{equation}
\label{eq:1}
\frac{C_1A_1}{A_1A_2}\cdot \frac{A_2Z}{ZB_2}\cdot \frac{B_2B_1}{B_1C_1}=-1,
\end{equation}
\begin{equation}
\label{eq:2}
\frac{A_2A_3}{A_3C_2}\cdot \frac{C_2B_3}{B_3B_2}\cdot \frac{B_2Z}{ZA_2}=-1.
\end{equation}
上面两个方程等号的左右两边分别相乘,得到
\begin{equation}
\label{eq:3}
\frac{C_1A_1}{A_1A_2}\cdot \frac{A_2A_3}{A_3C_2}\cdot
\frac{C_2B_3}{B_3B_2}\cdot \frac{B_2B_1}{B_1C_1}=1.
\end{equation}

\begin{equation}
\label{eq:4}
\left.\frac{C_1A_1}{A_1A_2}\middle/
\frac{C_1B_1}{B_1B_2}\right.=\left.\frac{C_2A_3}{A_3A_2}\middle/
\frac{C_2B_3}{B_3B_2}\right.\iff \left.\frac{A_1C_1}{A_1A_2}\middle/
\frac{B_1C_1}{B_1B_2}\right.=\left.\frac{A_3C_2}{A_3A_2}\middle/ \frac{B_3C_2}{B_3B_2}\right.
\end{equation}
然后取极限情形,让点$B_2$与点$A_2$重合,则图1会变成图2.
图2
相应的,式\eqref{eq:4}就会变成
\begin{equation}
\label{eq:5}
\left.\frac{A_1C_1}{A_1Z}\middle/
\frac{B_1C_1}{B_1Z}\right.=\left.\frac{A_3C_2}{A_3Z}\middle/ \frac{B_3C_2}{B_3Z}\right.
\end{equation}
此即交比在射影下不变.