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doc88.com/p-0079125671261.html
定理 任凸四边形 $A_{1} A_{2} A_{3} A_{4}$ 中, $\triangle A_{2} A_{3} A_{4}, \triangle A_{3} A_{4} A_{1}, \triangle A_{4} A_{1} A_{2}$, $\triangle A_{1} A_{2} A_3$ 的外接圆半径分别为 $R_{1}, R_{2}, R_{3}, R_{4}$ 则
$$
\begin{aligned}
&\left(R_{1} R_{2}+R_{3} R_{4}\right) A_{1} A_{2} \cdot A_{3} A_{4}+\left(R_{1} R_{4}+
R_{2} R_{3}\right) A_{1} A_{4} \cdot A_{2} A_{3} \\
&=\left(R_{1} R_3+R_{2} R_{4}\right) A_{1} A_{3} \cdot A_{2} A_{4}
\end{aligned}
$$
证明:
设对角线交于$O$,由斯特槐定理,
\begin{aligned} & A_{1} O^{2} \cdot A_{2} A_{4} = A_{1} A_{4}^{2} \cdot A_{2} O + A_{1} A_{2}^{2} \cdot O A_{4}-A_{2} O \cdot O A_{4} \cdot A_{2} A_{4} \end{aligned}
移项, 两边乘以$ \frac{A_{2} A_{4} \cdot O A_{3}}{O A_{2}^{2} \cdot A_{1} A_{3}} $, 得
\begin{align*}
&\frac{A_{1} A_{4}^{2} \cdot A_{2} A_{4} \cdot O A_{3}}{O A_{2} \cdot A_{1} A_{3}}\\
&=\frac{A_{1} O^{2} \cdot A_{2} A_{4}^{2} \cdot O A_{3}}{O A_{2}^{2} \cdot A_{1} A_{3}}\\
&+\frac{A_{2} A_{4}^{2} \cdot O A_{4} \cdot O A_{3}}{O A_{2} \cdot A_{1} A_{3}}\\
&-\frac{A_{1} A_{2}^{2} \cdot O A_{4} \cdot A_{2} A_{4} \cdot O A_{3}}{A_{2} O^{2} \cdot A_{1} A_{3}} \tag1
\end{align*}
同理\begin{aligned}
&\frac{A_{3} A_{4}^{2} \cdot A_{2} A_{4} \cdot O A_{1}}{O A_{2} \cdot A_{1} A_{3}}\\
&=\frac{O A_{3}^{2} \cdot A_{2} A_{4}^{2} \cdot O A_{1}}{O A_{2}^{2} \cdot A_{1} A_{3}}\\
&-\frac{A_{2} A_{3}^{2} \cdot A_{2} A_{4} \cdot O A_{1} \cdot O A_{4}}{O A_{2}^{2} \cdot A_{1} A_{3}}\\
&+\frac{A_{2} A_{4}^{2} \cdot O A_{1} \cdot O A_{4}}{O A_{2} \cdot A_{1} A_{3}} .
\end{aligned}
(1)+(2) ，应用
\begin{aligned}
&\text { } \\
&\frac{A_{2} A_{4}}{O A_{2}}=\frac{\triangle A_{2} A_{3} A_{4}}{\triangle A_{2} O A_{3}}, \\
&\frac{O A_{3}}{A A_{3}}=\frac{\triangle A_{2} O A_{3}}{\triangle A_{1} A_{2} A_{3}}\end{aligned}
即$$ \frac{A_{2} A_{4} \cdot O A_{3}}{O A_{2} \cdot {A}_{1} A_{3}}=\frac{\triangle A_{2} A_{3} A_{4}}{\triangle A_{1} A_{2} A_{3}}$$
以及 $\frac{A_{2} A_{4} \cdot O A_{1}}{O A_{2} \cdot A_{1} A_{3}}=\frac{\triangle A_{4} A_{1} A_{2}}{\triangle A_{1} A_{2} A_{3}}$,
$$
\frac{O A_{4}}{O A_{2}}=\frac{\triangle A_{3} A_{4} A_{1}}{\triangle A_{1} A_{2} A_{3}},
$$
再考虑
$$
\begin{aligned}
&\triangle A_{1} A_{2} A_{3}=\frac{A_{1} A_{2} \cdot A_{1} A_{3} \cdot A_{2} A_{3}}{4 R_{1}} \\
&\triangle A_{4} A_{2} A_{3}=\frac{A_{2} A_{3} \cdot A_{3} A_{4} \cdot A_{2} A_{4}}{4 R_{1}} \\
&\triangle A_{4} A_{1} A_{3}=\frac{A_{1} A_{3} \cdot A_{3} A_{4} \cdot A_{4} A_{1}}{4 R_{2}} \\
&\triangle A_{4} A_{1} A_{2}=\frac{A_{1} A_{2} \cdot A_{2} A_{4} \cdot A_{4} A_{1}}{4 R_{3}}
\end{aligned}
$$
即得欲证。
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