A set $S$ (in some Euclidean space) is called convex if, for every two vectors $a,b$ in $S$, every convex combination of $a,b$ is also in $S$, that is, $t a + (a-t)b \in S$ for all $t\in[0,1]$.
Is it equivalent to define a convex set as follows: for every two vectors $a,b$ in $S$, the arithmetic mean $a/2 + b/2$ is in $S$?
One direction is obvious. For the other direction, I thought that, if $a/2+b/2$ is in $S$ then we can apply the same definition to $a$ and $a/2+b/2$, and get that $3a/4+b/4$ is also in $S$; similarly, $a/4+3b/4 \in S$. Proceeding by induction, we can prove that, whenever $t = p/2^q$ for some integers $p,q$, we have $t a + (a-t)b \in S$. Is it possible to extend the proof to all $t$?
As others have said, this property is midpoint-convexity. The rationals $\mathbb{Q}$ are midpoint convex but not convex. However, any closed set that is midpoint convex is convex, which is a nice exercise and is along the lines of the argument you have written. You need closedness to extend from the dyadic rationals to all $t$.