回復 1# bch0722b 的帖子
很像蝴蝶定理
令
\(\begin{align}
& \angle AOE=\angle COF=\alpha \\
& \angle BOE=\angle DOF=\beta \\
& \angle BOH=\angle DOG=\gamma \\
& \angle COH=\angle AOG=\omega \\
\end{align}\)
由張角定理
\(\begin{align}
& \frac{1}{\overline{OE}}=\frac{\sin \beta }{\overline{OA}}+\frac{\sin \alpha }{\overline{OB}} \\
& \frac{1}{\overline{OH}}=\frac{\sin \omega }{\overline{OB}}+\frac{\sin \gamma }{\overline{OC}} \\
& \frac{\sin \left( \beta +\gamma \right)}{\overline{OI}}=\frac{\sin \gamma }{\overline{OE}}+\frac{\sin \beta }{\overline{OH}}=\frac{\sin \beta \sin \gamma }{\overline{OA}}+\frac{\sin \alpha \sin \gamma }{\overline{OB}}+\frac{\sin \beta \sin \omega }{\overline{OB}}+\frac{\sin \beta \sin \gamma }{\overline{OC}} \\
& \\
& \frac{1}{\overline{OF}}=\frac{\sin \beta }{\overline{OC}}+\frac{\sin \alpha }{\overline{OD}} \\
& \frac{1}{\overline{OG}}=\frac{\sin \gamma }{\overline{OA}}+\frac{\sin \omega }{\overline{OD}} \\
& \frac{\sin \left( \beta +\gamma \right)}{\overline{OJ}}=\frac{\sin \gamma }{\overline{OF}}+\frac{\sin \beta }{\overline{OG}}=\frac{\sin \beta \sin \gamma }{\overline{OC}}+\frac{\sin \alpha \sin \gamma }{\overline{OD}}+\frac{\sin \beta \sin \gamma }{\overline{OA}}+\frac{\sin \beta \sin \omega }{\overline{OD}} \\
& \\
& \overline{OB}=\overline{OD} \\
& \overline{OI}=\overline{OJ} \\
\end{align}\)
