Proof of the Mazur-Ulam Theorem The Mazur-Ulam Theorem Theorem $2.1$ states that any surjective isometry between any two real normed spaces $f:X \rightarrow Y$ is affine.
In the proof of the theorem, the author mentioned that it suffices to show that for any $x, y \in X$,
$$f\left(\dfrac{x+y}{2}\right) = \dfrac{f(x)+f(y)}{2}$$
Why is it the case? How to conclude $f$ is affine from equation above?
Recall that $f:X \rightarrow Y$ is an affine function if for all $x,y \in X$ and $0 \leq t \leq 1$,
$$f[(1-t)x+ty] = (1-t)f(x) + t f(y)$$
 A: If it is given that 
$$
f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}
$$
Then 
$$
f\left(\frac{x+y+z+w}{4}\right)=\frac{f(\frac{x+y}{2})+f(\frac{z+w}{2})}{2}
=\frac{f(x)+f(y)+f(z)+f(w)}{4}
$$
So by induction,
$$
f\left(\frac{1}{2^n}\sum_{i=1}^{2^n}{a_i}\right) = \frac{1}{2^n}\sum_{i=1}^{2^n}{f(a_i)}
$$
And any real number between 0 and 1 can be estimated by something over $2^n$ for large enough $n$. 
A: The midpoint-affine property $$f\left(\dfrac{x+y}{2}\right) = \dfrac{f(x)+f(y)}{2} \tag{1}$$ implies being affine under the assumption that $f$ is continuous (which it is, being an isometry). As stated, $(1)$ amounts to the case $t=1/2$ of 
$$f[(1-t)x+ty] = (1-t)f(x) + t f(y)\tag{2}$$
But applying $(1)$ again, the second time to $x$ and $(x+y)/2$, yields $(2)$ for $t=1/4$. Similarly, applying $(1)$ to $y$ and $(x+y)/2$ yields $(2)$ for $t=3/4$.
Continuing this process, we obtain $(2)$ for all dyadic rationals in $(0,1)$: numbers of the form $k/2^m$, $0<k<2^m$. These are dense in $[0,1]$ and since both sides of $(2)$ are continuous with respect to $t$, equality $(2)$ holds for all $t\in [0,1]$.
