回復 1# kuen 的帖子
設直徑上那五點由左至右分別是\({{O}_{2}},B,C,{{O}_{1}},A\)
易知\(\overline{BC}=2\)
作\(\overline{{{O}_{2}}D}\)垂直\(\overline{PQ}\)於\(D\)
易知\(\overline{PD}=\overline{QD}\)且\(\overline{{{O}_{2}}D}\)和\(\overline{PB}\)平行
\(\begin{align}
& \frac{\overline{AP}}{\overline{PD}}=\frac{\overline{AB}}{\overline{B{{O}_{2}}}}=\frac{12}{6}=2 \\
& \overline{AP}=2\overline{PD}=\overline{PQ} \\
\end{align}\)
