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# System prepended metadata

title: 證明題 2.
tags: [直線與圓]

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###### tags: `直線與圓`

# 證明題 2. 

己知 $\overline{PA}$、$\overline{PB}$ 為圓的兩切線，且切於 $A$，$B$ 兩點，過 $P$ 的直線交圓於 $C$、$D$ 兩點，與弦 $\overline{AB}$ 交於 $Q$，試證明：$\overline{PQ}^2=\overline{PC}\cdot\overline{PD}-\overline{QC}\cdot\overline{QD}$

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$\boxed{證}：$作一個簡圖
![證明題2.簡圖](https://i.imgur.com/Yl6wUjU.png =200x)
$\angle PQA=\pi-\angle PQB$
$\therefore \cos\angle PQA=-\cos\angle PQB$
$\begin{align}\frac{\overline{QA}^2+\overline{QP}^2-\overline{PA}^2}{2\cdot\overline{QA}\cdot\overline{QP}}=-\frac{\overline{QB}^2+\overline{QP}^2-\overline{PB}^2}{2\cdot\overline{QB}\cdot\overline{QP}}\end{align}$
$\begin{align}\overline{QB}(\overline{QA}^2+\overline{QP}^2-\overline{PA}^2)+\overline{QA}(\overline{QB}^2+\overline{QP}^2-\overline{PB}^2)=0\end{align}$
$\begin{align}(\overline{QA}+\overline{QB})(\overline{QA}\cdot\overline{QB}+\overline{QP}^2-\overline{PA}^2)=0\end{align}$
$\begin{align}\Rightarrow\overline{QA}\cdot\overline{QB}+\overline{QP}^2-\overline{PA}^2=0\end{align}$
$\begin{align}\overline{PQ}^2=\overline{PA}^2-\overline{QA}\cdot\overline{QB}=\overline{PC}\cdot\overline{PD}-\overline{QC}\cdot\overline{QD}\end{align}$