I think the following gives a proof, but I might have overlooked something.
Take two points $a,a'$ on $\alpha$, and let $b=f(a),b'=f(a')$. Apply an isometry of $\mathbb{R}^2$ to $\beta$, so as to get $a=b$ and $a'=b'$. Consider one of the segment from $a$ to $a'$ on $\alpha$ and name it $s$; then the points having same distance from $a$ and $a'$ than a point on $s$ form two arcs: $s$ itself and its mirror image with respect to $(aa')$. Due to convexity, one of the two segment between $a$ and $a'$ along $\beta$ must be equal to $s$. Proceed similarly for the other side.