I'm trying to understand if this way is correct always or not : 

$a,b,c$ natural numbers 

$b$ prime number 

$a,c$ even number

if $\gcd (a,b)=1$ 

and we have 

$c\equiv a\pmod{b}$ 

$\implies $ $\frac{c}{2}\equiv \frac{a}{2}\pmod{b}$ 

How I can understand and prove it ? Without using the Euclidean algorithm, because, I don't study $\frac{\mathbb{Z}}{p\mathbb{Z}}$

I know simple relations of modulo : 

$c=b.k+c$ then $\frac{c}{2}=b\frac{k}{2}+\frac{a}{2} %only if a=c$ 

But may be $\frac{k}{2}\not\in\mathbb{N}$ 

My goal is to solve this : 

$8n\equiv 1154\pmod{1163}$ 

So if we divide by $2$ we get 

$4n\equiv 577\pmod{1163}$

$$4n\equiv 1740\pmod{1163}$$ 
Then 

$$2n\equiv 870\pmod{1163}$$ 

Then 

$$n\equiv 435\pmod{1163}$$
 ( correct answer ) 

For example : 

$20\equiv 1\pmod{19}$ 

$20\equiv 20\pmod{19}$ 

Then $10\equiv 10\pmod{19}$

Correct 

So what's the proof ?