(a) Prove that $a|b$ if and only if $\gcd(a,b) = a$.
(b) Let $b > 9a$, Show that $\gcd(a,b) = \gcd(a,b−2a)$
(c) Show that If $a$ is even and $b$ is odd, then $\gcd(a,b) = \gcd(a/2,b)$
(d) Show that if $a$ is even and $b$ is even, then $\gcd(a,b) = 2\gcd(a/2, b/2)$
(Just started reading on this topic. I tried to solve them using the basic equation $gcd(a,b) = as + bt$ but I am not able to solve the above statements.)
$b)$. Let $d =\text{gcd}(a,b) \to d\mid a, d\mid b \to d\mid b-2a \to d\mid \text{gcd}(a,b-2a)$. Now if $d\mid \text{gcd}(a,b-2a) \to d\mid a, d\mid b-2a \to d\mid (2a + (b-2a)) = b \to d\mid \text{gcd}(a,b) \to \text{gcd}(a,b) = \text{gcd}(a,b-2a)$.
$c)$. Let $a = 2m \to \text{gcd}(a,b) = \text{gcd}(2m,b) = d \to d\mid 2m, d\mid b \to d \text{ is odd} \to d\mid m \to d\mid \text{gcd}(m,d) = \text{gcd}(a/2,b)$. Conversely if $d =\text{gcd}(a/2,b) \to d\mid a/2, d\mid b \to d\mid 2(a/2) = a \to d\mid \text{gcd}(a,b) \to \text{gcd}(a,b) = \text{gcd}(a/2,b)$.
$d)$. Let $a = 2m, b = 2n \to \text{gcd}(a,b) = \text{gcd}(2m,2n)$. If $d = 2\cdot \text{gcd}(m,n) \to d\mid 2m, d\mid 2n$ since $\text{gcd}(m,n)\mid m$, and $\text{gcd}(m,n)\mid n \to d\mid \text{gcd}(2m,2n) = \text{gcd}(a,b) \to 2\cdot \text{gcd}(a/2,b/2) \mid \text{gcd}(a,b)$. Conversely if $d= \text{gcd}(a,b) = \text{gcd}(2m,2n)$. For this particular case if $d$ is even, then $d = 2s \to 2s\mid 2m, 2s\mid 2n \to s\mid m, s\mid n \to s\mid \text{gcd}(m,n) \to d = 2s\mid 2\cdot \text{gcd}(m,n) = 2\cdot \text{gcd}(a/2,b/2)$. If $d$ is odd, then $d\mid 2m, d\mid 2n \to d\mid m, d\mid n \to d\mid \text{gcd}(m,n) = \text{gcd}(a/2,b/2) \to d\mid 2\cdot \text{gcd}(a/2,b/2) \to \text{gcd}(a,b) = 2\text{gcd}(a/2,b/2)$.