完全四边形

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目的是重画page19

首先尝试标记点

需要标注点和线的坐标

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标记了点坐标后，它已经变得有点混乱了

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I find the locus of the pole of a given line $\lambda \alpha+\mu \beta+\nu \gamma=0$, with respect to a conic which passes through four fixed points, $(f: \pm g: \pm h)$.

By proper choice of the reference triangle the conics of this pencil may be expressed by the equation $a \alpha^2+b\beta^2+c \gamma^2=0$, where $a, b$, and $c$, are constants having different values for each conic of the pencil.
The polar of any point $\left(\alpha_1: \beta_1: \gamma_1\right)$ will have the form
\begin{equation}
a \alpha \alpha_1+b\beta \beta_1+c \gamma \gamma_1=0
\end{equation}
Since the conics pass through $(f: \pm g: \pm h)$ we have the condition
\begin{equation}
a f^2+b g^2+c h^2=0
\end{equation}
The polar (1) is to be indentified with the line $\lambda \alpha+\mu \beta+\nu \gamma=0$, since $\left(\alpha_1: \beta_1: \gamma_1\right)$ is to be the pole of this line. Therefore, identifying coefficients,
\[
\begin{array}{lll}
\lambda=a \alpha_1 & \text { or } & a=\frac{\lambda}{\alpha_1} \\
\mu=b \beta_1 & \text { or } & b=\frac{\mu}{\beta_1} \\
\nu=c \gamma_1 & \text { or } & c=\frac{\nu}{\gamma_1}
\end{array}
\]Putting these values for $a, b$, and $c$, in (2) gives
\[\frac{\lambda f^2}{\alpha_1}+\frac{\mu g^2}{\beta_1}+\frac{\nu h^2}{\gamma_1}=0 .
\]
since $\left(\alpha_1: \beta_1: \gamma_1\right)$ was any point, by removing the subscripts and letting the point vary we have the locus:
\[
\frac{\lambda f^2}{\alpha}+\frac{\mu g^2}{\beta}+\frac{\nu h^2}{\gamma}=0, \quad \text { which is evidently a conic. }
\]