For odd natural numbers $a,b,c$, prove that: $$\gcd \left( \frac{a+b}{2}, \frac{b+c}{2}, \frac{a+c}{2} \right) = \gcd(a, b, c).$$
How can we deal with the fact that: $$a = \frac{a + c}{2} + \frac{a + b}{2} - \frac{b + c}{2}.$$
Well, we know that if b > a, then gcd(a, b) = gcd(a, b-a). Using this property I resulted $$\gcd \left( \frac{a+b}{2}, \frac{b+c}{2}, \frac{a+c}{2} \right)$$ in the form $$\gcd \left( b, \frac{b+c}{2}, \frac{a-b}{2} \right)$$ It seems to me that all that remains is to achieve a and c from the $$\frac{b+c}{2}, \frac{a-b}{2}$$ by performing arithmetic operations. All my attempts to do this were unsuccessful...
It seems to me that a slightly different approach is needed here. Therefore, if you have any ideas or suggestions for a solution, please write them, I will be grateful for any hint!
John Douma has basically answered the question in a comment.
The fact that you can express $a$, and analogously $b$ and $c$, as a sum and difference of $r=\frac{a+b}2$, $s=\frac{b+c}2$ and $t=\frac{a+c}2$ shows that any common divisor of $r$, $s$ and $t$ is also a common divisor of $a$, $b$ and $c$. For the other direction, since $a$, $b$ and $c$ are odd, their common divisor isn’t divisble by $2$, so e.g. $r=\frac{a+b}2$ is divisible by all common divisors of $a$ and $b$ since the $2$ in the denominator doesn’t divide any of them.
Since all common divisors on one side are also common divisors on the other side, the greatest common divisors coincide.