Hi,

try it this way:

d_1=gcd(c,b), d_2=gcd(ac,b)

=> cx_1+by_1=d_1 and acx_2+by_2=d_2

From gcd(a,b)=1 you get

=> ax+by=1 | *d_1

=> d_1=cx_1+by_1

Now we have to show, that d_2|d_1:

d_1*(ax+by=1) => ax(cx_1+by_1)+bd_1y=d_1 => ac(xx_1)+b(axy_1+d_1y)=d_1

You know d_2=gcd(ac,b), so d_2 divides any linear combination of ac and b => d_2|d_1

Repeat the steps for d_2 and you will get d_1|d_2 and d_2|d_1 => d_1=d_2

Greetings