Indices are taken $\pmod{5}$, as in the question. Finding the cross ratiosSuppose $A_{i-1}B_{i-1}$, $A_iB_i$, $A_{i+1}B_{i+1}$ are concurrent. We claim that $(B_i,A_{i+1};A_{i+2},B_{i-2})=(B_i,A_{i-1};A_{i-2},B_{i+2})$. Proof: By Desargues' on $\triangle A_iB_{i+1}1B_{i-1}$ and $\triangle B_iA_{i+1}A_{i-1}$, the lines $B_{i-2}B_{i+2},A_{i-1}A_{i+1},A_{i-2}A_{i+2}$ are concurrent, which implies the claim. Also, note that Desargues' on $\triangle A_iB_{i+2}1B_{i-2}$ and $\triangle B_iA_{i+2}A_{i-2}$ means that the same two cross ratios are equal iff $A_{i-2}B_{i-2}$, $A_iB_i$, $A_{i+2}B_{i+2}$ are concurrent. (Interestingly, this directly implies that if $4$ of the $A_iB_i$ are concurrent, the fifth also passes through their intersection.) Using themNow suppose WLOG that $A_iB_i$ are concurrent where $i=1,2,3,4$. Since $A_2B_2,A_3B_3,A_4B_4$ concurrent, then $(B_1,A_5;A_4,B_3)=(B_5,A_1;A_2,B_3)$. Since $A_1B_1,A_3B_3,A_4B_4$ concurrent, then $(B_1,A_5;A_4,B_3)=(B_1,A_2;A_3,B_4)$. Let this common cross ratio be $r$ and let $A_2B_2$ intersect $B_1B_3$ at $X$. Then $$r^2 = (B_5,A_1;A_2,B_3) \times (B_1,A_2;A_3,B_4) = (A_4,A_5;X,B_3) \times (B_1,X;A_4,A_5) = \frac{A_4X \times A_5B_3}{A_4B_3 \times A_5X} \times \frac{B_1A_4 \times XA_5}{B_1A_5 \times XA_4} = \frac{A_5B_3 \times B_1A_4}{A_4B_3 \times B_1A_5} = \frac{A_5B_3 \times B_1A_4}{A_1B_3 \times B_4A_5 + A_5B_3 \times B_1A_4} = \frac{ - (B_1,A_5;A_4,B_3)}{1 - (B_1,A_5;A_4,B_3)} = \frac{-r}{1-r}$$ Solving for $r$, we get $r^2-r-1=0$, which does not have rational roots. However, note that all the sides of the pentagon can be described by linear equations with rational coefficients, so all the $B_i$ have rational coordinates. This means that $r$ has to be rational, contradiction.