By LoS, we have $\frac{X_1A_3}{X_1A_4}=\frac{\sin A_4}{\sin A_3}$. Similar computations for the other $4$ ratios yields $\prod_{i=1}^{5} \frac{X_iA_{i+2}}{X_iA_{i+3}}=1$, implying the result.

Let $\angle X_3 A_5 A_1 = \angle X_2 A_5 A_4 = a$, $\angle X_3A_1A_5 = \angle X_4A_1A_2 =b $, $\angle A_1A_2X_4 = \angle A_3A_2X_5 = c$, $\angle A_2A_3X_5=\angle A_4A_3X_1=d$, and $\angle X_2A_4A_5=\angle X_1A_4A_3 = e$. Then, $$\frac{X_1A_3 \cdot X_2 A_4 \cdot A_3A_5 \cdot X_4A_1 \cdot X_5A_2}{X_1A_4 \cdot X_2 A_5 \cdot X_3A_1 \cdot X_4A_2 \cdot X_5A_3} = \frac{\sin (e)}{\sin (d)} \cdot \frac{\sin (a)}{\sin (e)} \cdot \frac{\sin (b)}{\sin (a)} \cdot \frac{\sin (c)}{\sin (b)} \cdot \frac{\sin (d)}{\sin (c)}=1$$as desired.

By Sine Rule, $\dfrac{X_n A_{n+2}}{X_n A_{n+3}} = \dfrac{sin \angle X_n A_{n+2} A_{n+3}}{sin \angle X_n A_{n+3} A_{n+2}}$ Let $\angle a_n = \angle X_{n+2} A_{n+4} A_{n+5} \forall n$ Let $\angle b_n = \angle X_{n+2} A_n A_{n+4} \forall n$ Note that $a_n = b_{n-1} = 180 - \angle A_n A_{n+1} A_{n+2}$ Therefore, \begin{align*} &\dfrac{\prod _{i=1} ^5 A_i A_{i+2}}{\prod _{i=1} ^5 A_i A_{i+3}}\\ &= \dfrac{sin a_1 sin a_2 sin a_3 sin a_4 sin a_5}{sin b_1 sin b_2 sin b_3 sin b_4 sin b_5}\\ &=1\\ \end{align*}$\square$

AopsUser101 wrote: Let $\angle X_3 A_5 A_1 = \angle X_2 A_5 A_4 = a$, $\angle X_3A_1A_5 = \angle X_4A_1A_2 =b $, $\angle A_1A_2X_4 = \angle A_3A_2X_5 = c$, $\angle A_2A_3X_5=\angle A_4A_3X_1=d$, and $\angle X_2A_4A_5=\angle X_1A_4A_3 = e.$ Next, draw a perpendicular line from $X_i,$ to $A_{i+2}A_{i+3}$ for $1 \le i \le 5,$ where indices are taken modulo 5. Let the length of the line drawn $X_i$ be $h_i.$ By the definition of sine, $\sin (a)=\frac{h_3}{X_3 A_5}$ which means $X_3A_5=\frac{h_3}{\sin(a)}$ Similar steps for all other segments of the form $X_i A_{i+2}$ yield that $\prod_{i=1}^{5} X_iA_{i+2}= \frac{\prod_{i=1}^{5} h_i}{\sin(a) \sin(b) \sin(c) \sin(d) \sin(e)}.$ Going back to the definition of sine, we get that $\sin (a)=\frac{h_2}{X_2A_5}$ which means $X_2A_5=\frac{h_2}{\sin(a)}.$ Once again, similar steps for all other segments of the form $X_iA_{i+3}$ yield that $\prod_{i=1}^{5} X_iA_{i+3}= \frac{\prod_{i=1}^{5} h_i}{\sin(a) \sin(b) \sin(c) \sin(d) \sin(e)}.$ Therefore we have shown that $$\displaystyle\prod_{i=1}^{5} X_iA_{i+2} = \displaystyle\prod_{i=1}^{5} X_iA_{i+3}.$$