1. Inversion and Isogonal Conjugate:
- Perform an inversion through point $P$. This inversion maps the cyclic $2n$-gon $A_1A_2 \ldots A_{2n}$ to another cyclic $2n$-gon.
- Let $Q$ be the isogonal conjugate of $P$ with respect to the $2n$-gon $A_1A_2 \ldots A_{2n}$.

2. Distance Notations:
- Let $|PA_i| = p_i$ and $|QA_i| = q_i$.
- Let $A_iA_{i+1} = m_i$.

3. Similarity of Triangles:
- Since $\angle PA_1A_2 = \angle PA_2A_3 = \ldots = \angle PA_{2n}A_1$, triangles $PA_{i+1}A_i$ and $QA_{i+1}A_{i+2}$ are similar.
- This similarity gives us the ratio:
$$\frac{m_{2i}}{m_{2i+1}} = \frac{q_{2i+1}}{p_{2i-1}} = \frac{q_{2i}}{p_{2i}}$$
- We can write this as:
$$\frac{m_{2i}}{m_{2i+1}} = \sqrt{\frac{q_{2i+1}}{p_{2i-1}} \cdot \frac{q_{2i}}{p_{2i}}}$$

4. Product of Ratios:
- Multiply all the ratios for $i = 1, 2, \ldots, n$:
$$\frac{\prod_{i=1}^{n} m_{2i}}{\prod_{i=0}^{n-1} m_{2i+1}} = \sqrt{\frac{\prod_{i=1}^{2n} q_i}{\prod_{i=1}^{2n} p_i}}$$

5. Symmetry Argument:
- By symmetry, we also have:
$$\frac{\prod_{i=1}^{n} m_{2i}}{\prod_{i=0}^{n-1} m_{2i+1}} = \sqrt{\frac{\prod_{i=1}^{2n} p_i}{\prod_{i=1}^{2n} q_i}}$$

6. Equality of Products:
- Equating the two expressions, we get:
$$\sqrt{\frac{\prod_{i=1}^{2n} q_i}{\prod_{i=1}^{2n} p_i}} = \sqrt{\frac{\prod_{i=1}^{2n} p_i}{\prod_{i=1}^{2n} q_i}} = 1$$
- This implies:
$$\frac{\prod_{i=1}^{n} m_{2i}}{\prod_{i=0}^{n-1} m_{2i+1}} = 1$$

7. Conclusion:
- Therefore, we have:
$$\prod_{i=1}^{n} m_{2i} = \prod_{i=0}^{n-1} m_{2i+1}$$

The final answer is $\boxed{ \prod_{i=1}^{n} \left|A_{2i - 1}A_{2i} \right| = \prod_{i=1}^{n} \left|A_{2i}A_{2i+1} \right| }$