I am looking for an elementary way to prove the following theorem.

>**Theorem.** Let $\alpha$ and $\beta$ be two simple convex curves in $\mathbb R^2$.
Assume 
$$\mathop{\rm length} \alpha=\mathop{\rm length}\beta$$
and there is a 1-Lipschitz bijecction $f\colon\alpha\to\beta$.
Then $f$ is an isometry.

It would be better if the same proof would work for Lobachevsky plane and unit sphere
(for the sphere one has to assume that the length of the curves is $<2{\cdot}\pi$).

The proof I know is simple, but it use Alexandrov geometry quite a bit:
*If we cut from the plane the region bounded by $\alpha$ 
and glue instead the region bounded by $\beta$
then the obtained space will have curvature $\ge0$ in the  sence of Alexandrov and 
it is easy to show that it has to be isometric to the Euclidean plane. Hence the result.*

**P.S.** This morning I realized that this also follows from a version of Cauchy's Arm Lemma;
so really simple solution is very unlikely. 

Namely, assume $\alpha,\beta\colon[0,\ell]\to\mathbb R^2$ are a closed convex curves
with unit-speed parameter.
Assume that for any $t$ curvature of $\alpha$ at $\alpha(t)$
is at most the curvature $\beta$ at $\beta(t)$
(In the non-smooth case the curvature is measure and this inequality still has sense).
Then $\alpha$ and $\beta$ are isometric.