Can you prove this implication between $\gcd$ equalities?

Question

Given six positive integers $\{a_1,b_1,a_2,b_2,a_3,b_3\}$ such that $$ \gcd(a_1,b_1) = \gcd(a_2,b_2) = \gcd(a_3,b_3) $$ then the following equality also holds, $$ \gcd(d_{31},d_{32}) = \gcd(d_{31},d_{21}) = \gcd(d_{32},d_{21}) $$ being $d_{ij}=|a_ib_j - a_jb_i| $.

Answer 1

By dividing through by the common gcd, it suffices to assume that all the input pairs are relatively prime. By symmetry, it suffices to show that, if $d \mid d_{31}$ and $d \mid d_{32}$, then $d \mid d_{21}$. We have that $d \mid a_2 d_{31} - a_1 d_{32} = a_3 d_{21}$ and $d \mid b_2 d_{31} - b_1 d_{32} = b_3 d_{21}$. Since $a_3$ and $b_3$ are relatively prime, this means that $d \mid d_{21}$.