A complete metric space with approximate midpoints is intrinsic Let $(X,d)$ be a metric space. For $x,y\in X$, define $A_{xy}$ the set of curves (the domain is supposed to be $[0,1]$) joining $x$ with $y$. For $\sigma\in A_{xy}$, define its length as $$L(\sigma)=\sup\sum_{i=1}^n d(\sigma(y_{i-1}),\sigma(y_i)),$$ where the supremum is taken over all partitions ${t_0,\ldots,t_n}$ of $[0,1]$.
Define $$d_L(x,y)=\inf_{\sigma\in A_{xy}} L(\sigma).$$
The metric $d$ is said to be intrinsic if $d=d_L$.
For $r>0$, $z$ is said to be $r-$midpoint for $x$ and $y$ if $2d(x,z)-d(x,y)<r$ and $2d(z,y)-d(x,y)<r$.
Prove that if $X$ is complete and for $x,y\in X$ and $r>0$ there exists a $r-$midpoint, then $d$ is intrinsic.
Given $r>0$, I tried to create a "curve" over the diadics in the interval $[0,1]$, sending a mid$-$point to a $r-$midpoint. Then extend the curve by continuity to $[0,1]$. I hoped this curve approximates arbitrary to the distance between its extremes.
Thanks for every hint or solution.
 A: Your idea is good, but $r$ should not stay the same for all dyadic points. As written, $r$ is an additive constant, and these can add up to a lot under the iterative construction. You'll need to use smaller $r$ as you go along. 
Let's begin by rewriting the assumed property of $X$ in a multiplicative way: for all $x,y\in X$ and all $t>1$ there is $z$ such that
$$
2\max(d(z,x)+d(z,y)) < t d(x,y) \tag{1}
$$
(This is obtained by letting $r=(t-1) d(x,y)$.)
Next, let $t_n$ be such that $\prod t_n$ converges and is close to $1$. Say, $t_n=\exp(\epsilon/n^2)$ where $\epsilon$ is small.
Then proceed as you wanted: define a function $f$ from the dyadic rationals on $[0,1]$ into $X$, starting  with $f(0)=x$, $f(1)=y$, and using the  midpoints as in (1) for the rest, with $t=t_n$ at the $n$th step. At $n$th step, the Lipschitz constant of $f$ grows at most by the factor $t_n$.   This implies that $f$ is Lipschitz continuous on dyadic rationals with the constant $\prod t_n$. Extend to $[0,1]$ by continuity; the Lipschitz constant is preserved.  This creates a curve of length at most $d(x,y)\prod t_n$, which can be arbitrarily close to $d(x,y)$. 
