Hilbert space satisfies the following condition: if two triangles $\triangle ABC$, $\triangle A_1B_1C_1$ have equal sides lengths: $AB=A_1B_1$, $BC=B_1C_1$, $AC=A_1C_1$ they also have equal medians lengths: $AM=A_1M_1$, where $M=(B+C)/2$, $M_1=(B_1+C_1)/2$. Are there other Banach spaces with this property? (If yes, there are 2dimensional examples, and the next question is complete classification. But I doubt.)
Firstly, notice that you may apply your condition to the triangles $ABM$ and $A_1B_1M_1$ and iterate this process. So, e.g.. for every points $X$ and $X_1$ on $BC$ and $B_1C_2$ with $BX/BC=B_1X_1/B_1C_1=k/2^n$ with integer $k,n$ we obtain $AX=A_1X_1$. Since the binary rationals are dense on $[0,1]$, the same holds for every ratio. Similarly, we may apply this to other sides and hence see that the affine transform mapping $ABC$ to $A_1B_1C_1$ is metricpreserving on the whole triangle $ABC$. THis already should suffice.
Indeed, consider an equilateral triangle $ABC$; by an affine transform we may assume that it is equilateral also in the Euclidean metric. Take a triangle $A'B'C'$ inscribed into $ABC$, with $A'$, $B'$, $C'$ dividing the sides of $ABC$ at the same ratio. Then $A'B'C'$ is also regular (in both metrics), and under the affine transform mapping $ABC$ to $A'B'C'$ the metric scales at the coefficient $\mu$ depending only on the ratio at which the vertices of $A'B'C'$ divide the sides of $ABC$.
If the sides of $A'B'C'$ are rotated at $\pi/n$ with respect to the sides of $ABC$ (angles are also Euclidean) then we may iterate inscribing such triangles $n$ times getting a triangle homothetic to $ABC$ with a known ratio. Since the metric on this triangle is $\mu^n$ times the initial one, we obtain that $\mu$ is the same for both metrics. Thus the ratios of segments rotated at $\pi/n$ is the same in our metric and Euclidean one. This is what we want.
For passing to an arbitrary space, see Anton Petrunin's answer...
In dimension 2, your condition implies existence of rotation by arbitrary small angle. Simply apply your condition recurcevely to a sequence of points on the unit circle $\dots,A_0,A_1,A_2,\dots$ such that the midpoint of $[A_{i1}A_{i+1}]$ lies on $[OA_i)$. The later implies that the metric is Euclidean.
Now in your space any 2dimesional subspace is Euclidean. In particular the parallelogram identity holds. The later implies that this is a Hibert space.

$\begingroup$ I am confused. What are triangles to which we apply it? $\endgroup$ – Fedor Petrov Sep 30 '15 at 13:26

$\begingroup$ @FedorPetrov $\triangle A_{i1}OA_{i+1}$. $\endgroup$ – Anton Petrunin Sep 30 '15 at 20:47

$\begingroup$ But why $A_{i1}A_{i+1}$ does not depend on $i$? $\endgroup$ – Fedor Petrov Sep 30 '15 at 21:19

$\begingroup$ @FedorPetrov, I was solving a more general problem (did not read the formulation carefully). I thought your condition holds only if $AB=AC$. This is likely the reason of your confusion. These equalities are easy to see, but it seems that you are satisfied with the other solution :) $\endgroup$ – Anton Petrunin Sep 30 '15 at 22:09