Congruent placements have rigidity operators of equal rank. Equality of all pairwise distances gives equality of the Gram matrices of the differences from a fixed vertex, hence a linear isometry of \R^d taking each p_v - p_w to q_v - q_w, and that isometry intertwines the two operators. Now let c be a continuous family of placements on a preconnected parameter space, each equivalent to c(t_0), with c(t_0) attaining \rho_d(K). The parameters at which c(t) is congruent to c(t_0) form a closed set, since congruence is the equality of the squared edge-length map of K at c(t) and at c(t_0) and that map is continuous along c. They form an open set as well: at such a parameter c(t) is congruent to c(t_0), so R_G has rank \rho_d(K) there too and G is locally rigid there by the previous proof; every nearby c(s) is equivalent to c(t) and close to it, hence congruent to c(t) and so to c(t_0). The set is nonempty and the parameter space is preconnected, so the set is everything, and taking the unit interval for the parameter space gives continuous rigidity at every placement of rank \rho_d(K).

For the converse at a regular placement, the slice of the previous proof gives paths inside the level set of the squared edge-length map of G whose tangent directions at p are all of \ker R_G(p). Continuous rigidity makes every pairwise distance constant along each of them, and differentiating at the base point gives \ker R_G(p) \subseteq \ker R_K(p) again. So continuous and local rigidity agree at every placement regular for both graphs, and the two open dense formulations agree with each other and with maximum rank.