Let \(x\) be an infinitesimal motion of \(G'\) at \(\hat p\), and let \(\rho (x)\) be its restriction to \(V\). Every edge \(uw\) of \(G\) is an edge of \(G'\). The entry of the rigidity map at \(uw\) depends only on the points and the velocities of \(u\) and \(w\) (lemma 6.4). So \(\rho (x)\) is an infinitesimal motion of \(G\) at \(p\), and \(\rho \) is a linear map from \(\ker R(G', \hat p)\) to \(\ker R(G, p)\).

We show that \(\rho \) is injective. Suppose that \(\rho (x) = 0\), so that \(x_u = 0\) for every \(u \in V\). The entries of \(R(G', \hat p)(x)\) at the new edges \(va\) and \(vb\) vanish:

Since \(x_a = x_b = 0\), the velocity \(x_v\) is orthogonal to \(q - p_a\) and to \(q - p_b\). These two vectors are linearly independent, so they span \({\mathbb {R}}^{2}\), and hence \(x_v = 0\). So \(x = 0\).

Since \(\rho \) is injective and \(p\) is infinitesimally rigid for \(G\),

Let \(p\) be an injective framework on \(V\) in \({\mathbb {R}}^{2}\) that is infinitesimally rigid for \(G\). Since \(p\) is injective and \(a \ne b\), the points \(p_a\) and \(p_b\) are distinct. The line through them and the finitely many points \(p_u\), for \(u \in V\), do not cover \({\mathbb {R}}^{2}\). So we can choose a point \(q\) off this line that is none of the points \(p_u\). Since \(q\) is off the line through \(p_a\) and \(p_b\), the vectors \(q - p_a\) and \(q - p_b\) are linearly independent.

Let \(\hat p\) be the framework that extends \(p\) by placing the new vertex at \(q\). It is infinitesimally rigid for \(G'\) (theorem 7.1). It is injective, since \(p\) is injective and \(q\) is none of the points \(p_u\).

As in the proof of theorem 7.1, let \(\rho (x)\) be the restriction to \(V\) of an infinitesimal motion \(x\) of \(G'\) at \(\hat p\). We first show that \(\rho (x)\) is an infinitesimal motion of \(G\) at \(p\). Every edge of \(G\) other than \(ab\) is an edge of \(G'\), so \(\rho (x)\) satisfies the constraint of each such edge, as before.

If \(ab\) is an edge of \(G\), the move deletes it, and we recover its constraint from the new edges \(va\) and \(vb\). We have \(q - p_a = \alpha (p_b - p_a)\) and \(q - p_b = (\alpha - 1)(p_b - p_a)\). The entries of \(R(G', \hat p)(x)\) at \(va\) and \(vb\) vanish (lemma 6.4). Dividing them by \(\alpha \) and by \(\alpha - 1\), which are not zero, gives

Subtracting the first equation from the second gives \(\langle p_b - p_a,\, x_a - x_b \rangle = 0\). This is the constraint \(\langle p_a - p_b,\, x_a - x_b \rangle = 0\) of the edge \(ab\), up to sign.

It remains to show that \(\rho \) is injective. If \(\rho (x) = 0\), the entries at the new edges \(va\) and \(vc\) show that \(x_v\) is orthogonal to \(q - p_a\) and to \(q - p_c\). These vectors are linearly independent, so \(x_v = 0\) and \(x = 0\). As in the proof of theorem 7.1, it follows that \(\dim \ker R(G', \hat p) \le \dim \ker R(G, p) \le 3\).

For \(\alpha \in {\mathbb {R}}\), let \(q_\alpha = p_a + \alpha (p_b - p_a)\). Since \(p_b - p_a \ne 0\), different values of \(\alpha \) give different points \(q_\alpha \). So all but finitely many \(\alpha \) satisfy \(\alpha \ne 0, 1\) and \(q_\alpha \ne p_u\) for every \(u \in V\). Choose such an \(\alpha \), and let \(q = q_\alpha \). The vectors \(q - p_a = \alpha (p_b - p_a)\) and \(q - p_c = \alpha (p_b - p_a) - (p_c - p_a)\) are linearly independent, since \(p_b - p_a\) and \(p_c - p_a\) are and \(\alpha \ne 0\).

Let \(\hat p\) be the framework that extends \(p\) by placing the new vertex at \(q\). It is infinitesimally rigid for \(G'\) (theorem 7.3). It is injective, since \(p\) is injective and \(q\) is none of the points \(p_u\).

Let \(p\) be an injective framework on \(V\) in \({\mathbb {R}}^{2}\) that is infinitesimally rigid for \(G\). If \(p_b - p_a\) and \(p_c - p_a\) are linearly independent, then theorem 7.4 applies to \(p\).

Otherwise, since \(p_b \ne p_a\) (as \(p\) is injective and \(a \ne b\)), we have \(p_c - p_a = \delta (p_b - p_a)\) for some \(\delta \in {\mathbb {R}}\). Choose a vector \(w \in {\mathbb {R}}^{2}\) such that \(w\) and \(p_b - p_a\) are linearly independent. For \(t \in {\mathbb {R}}\), let \(p^t\) be the framework that agrees with \(p\) except at \(c\), where \(p^t_c = p_c + t w\). The map \(t \mapsto p^t\) is continuous, and \(p^0 = p\). Infinitesimal rigidity is an open condition on the framework (lemma 6.17). So is injectivity, since \(V\) is finite. Hence \(p^t\) is injective and infinitesimally rigid for \(G\) for every \(t\) close enough to \(0\).

Since \(c\) is neither \(a\) nor \(b\), the points of \(a\) and \(b\) do not move, and

For \(t \ne 0\) these two vectors are linearly independent, because \(w\) and \(p_b - p_a\) are. Choose such a \(t \ne 0\) close enough to \(0\), and apply theorem 7.4 to \(p^t\).