I proved the case n=2 and made a futile 3-hour attempt to generalise it. This is not a very hard problem but I had a very hard time solving it.

Very slick. Extend $PA_{i}$ to meet the circle again at $B_{i+1} \ne A_{i},$ now the key observation is that $B_{1} \dots B_{2n}$ is a rotated version of $A_1 \cdots A_{2n}$ about the center of the circle. If we let $PA_{i} \times PB_{i+1}$ (constant over all $i$) equal $r^2,$ the negative inversion at $P$ with center $r$ sends $A_1, A_2, \cdots A_{2n-1}, A_{2n}$ to $B_{2}, B_{3} \dots B_{2n}, B_{1}$ respectively. Inversion distance formula yields the desired equality. $\blacksquare$

这么弱智的题都可以放在TST Relabel $P$ as $Br^+$ for convenience. We first prove that there exists isogonal conj of $Br^+$ wrt $A_1 \ldots A_{2n}$, and we call it $Br^-$. Lemma: In quad $ABCD$, $AD$ cut $(BCD)$ at $E$, $\angle ABF=\angle BCF=\angle CDF=\angle DEF$, then there exists isogonal conj of $F$ wrt $ABCD$. This is easy to prove by angle chasing: since $ABEF$ concyclic and $(FCD)$ is tangent to $BC$, we just want $\angle AFB+\angle CFD=180$. But $\angle AFB+\angle CFD=\angle AEB+180-\angle BCD=180$. Done. Now invert $A_1 \ldots A_{2n}$ wrt $Br^-$ at any radius to $A'_1 \ldots A'_{2n}$ and combining concyclic it is easy to prove $A_1A_3 \ldots A_{2n-1}$ is similar to $A'_2A'_4 \ldots A'_{2n}$ and $A_2A_4 \ldots A_{2n}$ is similar to $A'_3A'_5 \ldots A'_{2n-1}A'_1$. Take $Br^+$ and $Br^-$ into account and we can get $A_1 \ldots A_{2n} and Br^+$ is similar to $A'_2 \ldots A'_{2n}A'_1 and Br^-$. So $\frac{\prod_{i=1}^{n} \left|A_{2i - 1}A_{2i} \right|}{\prod_{i=1}^{n} \left|A_{2i}A_{2i+1} \right|}=\frac{\prod_{i=1}^{n} \left|A'_{2i - 1}A'_{2i} \right|}{\prod_{i=1}^{n} \left|A'_{2i}A'_{2i+1} \right|}=\frac{\prod_{i=1}^{n} \left|A_{2i - 2}A_{2i - 1} \right|}{\prod_{i=1}^{n} \left|A_{2i - 1}A_{2i} \right|}=\frac{\prod_{i=1}^{n} \left|A_{2i}A_{2i+1} \right|}{\prod_{i=1}^{n} \left|A_{2i - 1}A_{2i} \right|}$, and we can get $\prod_{i=1}^{n} \left|A_{2i - 1}A_{2i} \right| = \prod_{i=1}^{n} \left|A_{2i}A_{2i+1} \right|$. Done!

LoloChen wrote: 这么弱智的题都可以放在TST 定理 【LoloChen】所有奥数几何题都是trivial。 证明：如果不会做题目就问LoloChen

笑死了 Actually invert wrt P suffices, which is just #3

Awesome problem! I realised my solution was a bit different from the ones posted above, so here it is: By inversion through $P$ as above we find $P$ has an isogonal conjugate with respect to $A_1A_2\dots A_{2n}$ - let this be $Q$. Now let $|PA_i|=p_i$ and $|QA_i|=q_i$. Also let $A_iA_{i+1}=m_i$. Now triangles $PA_{i+1}A_i$ and $QA_{i+1}A_{i+2}$ are similar, so $\frac{m_{2i}}{m_{2i+1}}=\frac{q_{2i+1}}{p_{2i-1}}=\frac{q_{2i}}{p_{2i}}$ and we write this as $\frac{m_{2i}}{m_{2i+1}}=\sqrt{\frac{q_{2i+1}}{p_{2i-1}}\frac{q_{2i}}{p_{2i}}}$. By multiplying all of the above we get $\frac{\prod_{i=1}^{n} m_{2i}}{\prod_{i=0}^{n-1} m_{2i+1}}=\sqrt{\frac{\prod_{i=1}^{2n} q_i}{\prod_{i=1}^{2n} p_i}}$. However, by symmetry we also get $\frac{\prod_{i=1}^{n} m_{2i}}{\prod_{i=0}^{n-1} m_{2i+1}}=\sqrt{\frac{\prod_{i=1}^{2n} p_i}{\prod_{i=1}^{2n} q_i}}$, so $\sqrt{\frac{\prod_{i=1}^{2n} q_i}{\prod_{i=1}^{2n} p_i}}=\sqrt{\frac{\prod_{i=1}^{2n} p_i}{\prod_{i=1}^{2n} q_i}}=1$, so $\frac{\prod_{i=1}^{n} m_{2i}}{\prod_{i=0}^{n-1} m_{2i+1}}=1$. $\square$

I solved by complex numbers, but I'm not allowed to post it here (equations!), because I'm a new user. I'm surprised!

Suppose the polygon is inscribed in the unit circle. Let be $z_i\in C$, $I=1,2,...,2n$, each one of the vertex $A_i$ and the point $P$ denoted by $p\in C$. Also, $\angle PA_iA_{i+1}=\alpha$. Rotating vector $A_iP$ around $A_i$ clockwise, we have $$ \dfrac{z_{i+1}-z_i}{(p-z_i)\cdot cis\,\alpha}\in R$$$$ \therefore\dfrac{z_{i+1}-z_i}{(p-z_i)\cdot cis\,\alpha}=\dfrac{\overline{z_{i+1}}-\overline{z_i}}{(\overline{p}-\overline{z_i})\cdot \overline{cis\,\alpha}}=\dfrac{\dfrac{1}{z_{i+1}}-\dfrac{1}{z_{i}}}{(\overline{p}-\dfrac{1}{z_i})\cdot \dfrac{1}{cis\,\alpha}}=\dfrac{(z_i-z_{i+1})cis\,\alpha}{z_{i+1}(z_i\overline{p}-1)},$$$$\therefore z_{i+1}=\dfrac{p-z_i}{1-z_i\overline{p}}\cdot cis(2\alpha),\,\,\,\, \text{for}\,i=1,2,...,2n.$$ From this equality we get $$z_{i+2}-z_{i+1}=\left(\dfrac{p-z_{i+1}}{1-z_{i+1}\overline{p}}-\dfrac{p-z_i}{1-z_i\overline{p}} \right)\cdot cis(2\alpha)$$$$\therefore |z_{i+2}-z_{i+1}|=\left|\dfrac{p-z_{i+1}}{1-z_{i+1}\overline{p}}-\dfrac{p-z_i}{1-z_i\overline{p}} \right| =\dfrac{|(p-z_{i+1})(1-z_i\overline{p})-(p-z_i)(1-z_{i+1}\overline{p})|}{|1-z_{i+1}\overline{p}|\cdot|1-z_i\overline{p}|}\\ =\dfrac{|z_i-z_{i+1}|\cdot|1-p\overline{p}|}{|1-z_{i+1}\overline{p}|\cdot|1-z_i\overline{p}|}$$$$\therefore\dfrac{|A_{i+1}A_{i+2}|}{|A_iA_{i+1}|}=\dfrac{1-|p|^2}{|1-z_{i+1}\overline{p}|\cdot|1-z_i\overline{p}|}$$ But $|1-z_i\overline{p}|=|z_i(\overline{z_i}-\overline{p})|=|z_i-p|$. So, we rewrite the last equality above as $$\therefore\dfrac{|A_{i+1}A_{i+2}|}{|A_iA_{i+1}|}=\dfrac{1-|p|^2}{|z_{i+1}-p|\cdot|z_i-p|},$$and then $$1=\prod_{I=1}^{2n}\dfrac{|A_{i+1}A_{i+2}|}{|A_iA_{i+1}|}=\dfrac{(1-|p|^2)^{2n}}{|z_1-p|^2\cdot|z_2-p|^2\cdot...\cdot|z_{2n}-p|^2}$$$$\therefore |z_1-p|\cdot|z_2-p|\cdot...\cdot|z_{2n}-p|=(1-|p|^2)^{n}.$$Finally, $$\prod_{i=1}^{n}\dfrac{|A_{2i}A_{2i+1}|}{|A_{2i-1}A_{2i}|}=\dfrac{(1-|p|^2)^{n}}{|z_1-p|\cdot|z_2-p|\cdot...\cdot|z_{2n}-p|}=1,$$and we are done.

a22886 wrote: 笑死了 Actually invert wrt P suffices, which is just #3 why inversion suffices, how to show that the inverted polygon is similar to the original one, equal angles do not imply similar polygon.

Let $O$ be the circumcenter of $A_1A_2... A_{2n}$ and $O_i$ be the circumcenter of $\triangle PA_{i-1}A_{i}$. Claim 1: $\triangle PO_1O_2 \sim \triangle PA_1A_2$ Easy angle chasing shows that $\angle PO_1O_2 = \angle PA_4A_1 = \angle PA_1A_2$ and $\angle PO_2O_1 = \angle PA_2A_1$. Hence $\triangle PO_1O_2 \sim \triangle PA_1A_2$. So on so forth, we have the polygon $O_1O_2... O_{2n}$ is similar to $A_1A_2 ... A_{2n}$. Claim 2: $\triangle OO_3O_2 \sim \triangle PA_1A_2$ . Since $O_2O_3 \perp PA_2$ and $OO_2 \perp A_1A_2$, we have $\angle O2O_3O = \angle PA_2A_3 = \angle PA_1A_2$, similarly $\angle O_3O_2O = \angle PA_2A_1$, hence $\triangle OO_3O_2 \sim \triangle PA_1A_2$. On one hand, \[ \frac{A_1A_2}{A_2A_3} = \frac{O_1O_2}{O_2O_3} \] On the other hand, \[ \frac{A_1A_2}{A_2A_3} = \frac{A_1A_2}{PA_2} \cdot \frac{PA_2}{A_2A_3} = \frac{O_2O_3}{OO_2} \cdot \frac{OO_4}{O_3O_4}\] Multiplying them cyclicly, we have \[ \prod (\frac{A_{2i-1}A_{2i}}{A_{2i}A_{2i+1}}) ^2 = \prod \frac{O_{2i-1}O_{2i}}{O_{2i}O_{2i+1}} \times \prod \frac{O_{2i}O_{2i+1}}{O_{2i+1}O_{2i+2}} \times \prod \frac{OO_{2i}}{OO_{2i+2}} = 1 \] Remark: from claim 2, we have $O$ is isogonal conjugate to $P$ with respect to $O_1O_2... O_{2n}$.

Extremely Nice Problem. Extend $PA_i$ to meet the circumcircle at $B_i$, notice that the $B_i$ are basically rotations of $A_i$, Use inversion distance formula to complete.

We use complex numbers. Let $a_1$, $a_2$, $\dots$, $a_{2n}$ be on the unit circle in counterclockwise order and $z$ be a complex number for $P$. Let $\angle PA_1A_2=\theta$. We have \begin{align*}e^{i\theta}\cdot \frac{a_{i+1}-a_{i}}{z-a_i} &= \overline{e^{i\theta}\cdot \frac{a_{i+1}-a_{i}}{z-a_i}} \\ e^{i\theta}\cdot \frac{a_{i+1}-a_{i}}{z-a_i} &= e^{-i\theta}\cdot \frac{\overline{a_{i+1}}-\overline{a_{i}}}{\overline{z}-\overline{a_i}} \\ e^{2i\theta}(a_{i+1}-a_{i})\cdot \frac{\overline{z}-\overline{a_i}}{z-a_i} &= \overline{a_{i+1}}-\overline{a_{i}} \\ e^{2i\theta}(a_{i+1}-a_{i})\cdot \frac{\overline{z}-\overline{a_i}}{z-a_i} &= -(a_{i+1}-a_{i})(\overline{a_i}\overline{a_{i+1}}) \\ -e^{-2i\theta}\cdot \frac{z-a_i}{a_i\overline{z}-1} &= a_{i+1} \\ \end{align*}Note that \begin{align*} a_{i+1}-a_{i} &= -e^{-2i\theta}\left(\frac{z-a_i}{a_i\overline{z}-1}-\frac{z-a_{i-1}}{a_{i-1}\overline{z}-1}\right) \\ &= -e^{-2i\theta}\cdot \frac{(a_i-a_{i-1})(1-|z|^2)}{(a_i\overline{z}-1)(a_{i-1}\overline{z}-1)} \end{align*}Denote \begin{align*} P_1 &= \prod_{i=1}^{n} \left|A_{2i-1}A_{2i}\right| = \prod_{i=1}^{n} \left|a_{2i}-a_{2i-1}\right| \\ P_2 &= \prod_{i=1}^{n} \left|A_{2i}A_{2i+1}\right| = \prod_{i=1}^{n} \left|a_{2i+1}-a_{2i}\right|\\ P_3 &= \left|\left(1-|z|^2\right)^n\cdot \left(\prod_{i=1}^{2n} (a_i\overline{z}-1)\right)^{-1}\right| \end{align*}It's not hard to see that $P_1 = P_3P_2$ and $P_2=P_3P_1$ which proves that $P_1=P_2$.