prove $a c^{-1} \equiv b d^{-1} \bmod m$

Question

if $a \equiv b \bmod m$ and $c \equiv d \bmod m$ and $gcd(c,m) = 1$ prove that $a c^{-1} \equiv b d^{-1} \bmod m$ .

Answer 1

If $\gcd(c,m)=1$, then there exists $c^{-1}$ such that $cc^{-1}=1\pmod{m}$. Since $c\equiv{d}\pmod{m}$, $\gcd(d,m)=1$ and $dc^{-1}=1\pmod{m}$, and it's clear that $c^{-1}=d^{-1}\pmod{m}$. Then, it's obvious that $ac^{-1}=bd^{-1}\pmod{m}$.