Let $A_1A_2A_3A_4A_5A_6$ a hexagon and $A_6 A_2 \cap A_3 A_1=B_1$ , $A_1 A_3 \cap A_2 A_4=B_2$, $ A_2 A_4 \cap A_3 A_5=B_3,$, $A_3 A_5\cap A_4 A_6=B_4$, $A_4 A_6\cap A_5 A_1=B_5$, $A_5 A_1\cap A_6 A_2=B_6$.
If $$\frac {B_2B_3}{A_2A_4}=\frac {B_4B_5}{A_4A_6}= \frac {B_6B_1}{A_6A_2}$$ show that $$\frac {B_1B_2}{A_1A_3}=\frac {B_3B_4}{A_3A_5}=\frac {B_5B_6}{A_5A_1} .$$
I tried to prove it withs areas of triangles.
Hint: The figure below is not to scale:
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Note ($B_3B_4B_5B_5'$ is a parallelogram): $$\frac {B_2B_3}{A_2A_4}=\frac {B_4B_5}{A_4A_6} \Rightarrow \\ \frac {A_2A_4}{A_4A_6}=\frac {B_2B_3}{B_4B_5}=\frac {B_2B_3}{B_3B_5'}, \angle B_2B_3B_5'=\angle A_2A_4A_6 \Rightarrow \\ \Delta A_2A_4A_6\sim \Delta B_2B_3B_5' \Rightarrow \\ B_2B_5'||B_1B_6.$$