On the distance sets of AD-regular sets

I prove that if $\mathrm{\varnothing }\mathrm{\ne }K\subset {\mathbb{R}}^2$ is a compact $s$-Ahlfors-David regular set with $s\ge 1$, then $${\dim }_{\mathrm{p}}D(K)=1,$$ where $D(K):=\{\mathrm{\mid }x-y\mathrm{\mid }:x,y\in K\}$ is the distance set of $K$, and ${\dim }_{\mathrm{p}}$ stands for packing dimension. The same proof strategy applies to other problems of similar nature. For instance, one can show that if $\mathrm{\varnothing }\mathrm{\ne }K\subset {\mathbb{R}}^2$ is a compact $s$-Ahlfors-David regular set with $s\ge 1$, then there exists a point $x_0\in K$ such that ${\dim }_{\mathrm{p}}K\cdot (K-x_0)=1$. Specialising to product sets, one derives the following sum-product corollary: if $A\subset \mathbb{R}$ is a non-empty compact $s$-Ahlfors-David regular set with $s\ge 1\mathrm{/}2$, then $${\dim }_{\mathrm{p}}[A(A-a_1)+A(A-a_1)]=1$$ for some $a_1,a_2\in A$. In particular, $\dim_{$\mathrm{p}$} [AA + AA - AA - AA] = 1$. In all of the results mentioned above, compactness can be relaxed to boundedness and ${\mathcal{H}}^s$-measurability, if packing dimension is replaced by upper box dimension.