13. On a unit circle with center $O$, take 5 points in sequence $A_{1}, A_{2}, \cdots, A_{5} ; P$ is a point inside the circle. Denote the intersection of line segments $A_{i} A_{i+2}$ and $P A_{i+1}$ as $Q_{i}(i=1,2$, $\cdots, 5)$, where $A_{6}=A_{1}, A_{1}=A_{2}, O Q_{i}=d_{i}(i=1,2, \cdots, 5)$, try to find the product $A_{1} Q_{1} \cdot A_{2} Q_{2} \cdot \cdots \cdot A_{5} Q_{5}$.
13. Connect $A_{1} A_{2}, A_{2} A_{3}, A_{3} A_{4}, A_{4} A_{5}, A_{5} A_{1}$ and draw the diameter $M N$ through $Q_{1}$, as shown in Figure 12-21. By the intersecting chords theorem, we have $A_{1} Q_{1} \cdot Q_{1} A_{3}=M Q_{1} \cdot Q_{1} N=1-d_{1}^{2}$. Similarly, we have $A_{i} Q_{i} \cdot Q_{i} A_{i+2}=1-d_{i}^{2}, i=2,3,4,5$. Therefore, $\prod_{i=1}^{5}\left(A_{i} Q_{i} \cdot Q_{i} A_{i+2}\right)=\prod_{i=1}^{5}\left(1-d_{i}^{2}\right)$.
Since $\frac{A_{i} Q_{i}}{Q_{i} A_{i+2}}=\frac{S_{\triangle P A_{i} A_{i+1}}}{S_{\triangle P A_{i+1} A_{i+2}}}(i=1,2, \cdots, 5)$, we have $\prod_{i=5}^{5} \frac{A_{i} Q_{i}}{Q_{i} A_{i+2}}=1$. Combining these results, we get $A_{1} Q_{1} \cdot A_{2} Q_{2} \cdot \cdots \cdot A_{5} Q_{5}=\left[\prod_{i=1}^{5}\left(1-d_{i}^{2}\right)\right]^{\frac{1}{2}}$.