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

Theorem.Let $\alpha$ and $\beta$ be two simple convex closed 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 the following continuos version of Cauchy's Arm Lemma:

Let $\alpha,\beta\colon[0,\ell]\to\mathbb R^2$ be closed convex curves with unit-speed parameter. Assume that for any $t$ in a subinterval $[a,b]\subset [0,\ell]$, the curvature of $\alpha$ at $\alpha(t)$ is at most the curvature $\beta$ at $\beta(t)$. Then $|\alpha(a)-\alpha(b)|\ge|\beta(a)-\beta(b)|$ and equality holds only if the resriction $\alpha|[a,b]$ is isometric to the resriction $\beta|[a,b]$.