Apply a "relativization" of the procedure. The idea is simple, just the formalization is a tad delicate. Denote by $S_k$ the set obtained by a clockwise rotation of a set $S$ about the centre of the unit circle for $\dfrac {k\pi}{m}$ ($k=1,2,\ldots,2m$), thus $S_{2m} = S$. Let $B$ be made of $\beta$ disjoint arcs, each of length $\dfrac {\pi}{m}$ (thus with $1\leq \beta \leq 2m$; the case $\beta=0$, i.e. $B = \emptyset$, is trivial), so $\ell(B) = \dfrac {\beta\pi}{m}$. For an arc $b$ from $B$ notice that $\ell(b_i \cap b_j) = 0$ for $1\leq i < j \leq 2m$, thus $\displaystyle \ell\left (\bigcup_{k=1}^{2m} b_k\right ) = \sum_{k=1}^{2m} \ell(b_k) = 2\pi$. It means that the union of all $B_k$, $1\leq k \leq 2m$, covers each point on the unit circle (with the exception of maybe finitely many) precisely $\beta$ times. So the union of all $A\cap B_k$, $1\leq k \leq 2m$, covers each point on $A$ (with the exception of maybe finitely many) precisely $\beta$ times. That means there exists some $1\leq k \leq 2m$ such that $\ell(A^{2m-k}\cap B) = \ell(A\cap B_k) \geq \beta \dfrac {\ell(A)}{2m} = \dfrac {1}{2\pi}\ell(A)\dfrac {\beta \pi}{m} = \dfrac {1}{2\pi}\ell(A)\ell(B)$.