## 7 Proofs of lemmas in Section 2
**Lemma 11**: _For any facet \(F\) of \(V_{p}\), if \(H_{F}\) intersects \(B\cap\sigma\), then \(\angle(\mathbf{n}_{p},H_{F})\leqslant\arcsin(1/16)\) and both \(H_{F}\cap B\cap\sigma\) and \(F\cap B\cap\sigma\) contain no closed curve._  
Let \(pq\) be the dual Delaunay edge of \(F\) where \(H_{F}\) intersects \(pq\) at \(x\). We have \(\hat{d}(p,x)=\hat{d}(q,x)\leqslant\sqrt{4\lambda^{2}-w_{p}^{2}}\) from which we can derive that \(d(q,x)\leqslant\sqrt{4\lambda^{2}-w_{p}^{2}+w_{q}^{2}}\leqslant\sqrt{5}\lambda\). It follows that  
\[d(p,q)\leqslant d(p,x)+d(q,x)\leqslant(2+\sqrt{5})\lambda\leqslant 5\lambda.\]  
which is less than \(\frac{1}{8}\rho_{\sigma}\). By Lemma 10(ib), \(\angle(\mathbf{n}_{p},H_{F})=\pi/2-\angle(\mathbf{n}_{p},pq)\leqslant\arcsin(1/16)\). There is no closed curve in \(H_{F}\cap B\cap\sigma\) because such a closed curve would bound a 2-disk in \(B\cap\sigma\), which would contain a point \(x\) such that \(\angle(\mathbf{n}_{x},H_{F})=\pi/2\). This is a contradiction because \(\angle(\mathbf{n}_{x},H_{F})\leqslant\angle(\mathbf{n}_{x},\mathbf{n}_{p})+ \angle(\mathbf{n}_{p},H_{F})\leqslant 1/8+\arcsin(1/16)<\pi/2\) by Lemma 10(i.a). Since \(H_{F}\cap B\cap\sigma\) contains no closed curve, neither does \(F\cap B\cap\sigma\). This proves the claim. **Lemma 12**: _For any facet \(F\) of \(V_{p}\), if \(d(p,F)\leqslant\lambda\) and conditions C1-C3 hold, then \(F\cap B\cap\sigma\) is either empty or a single open curve._  
We assume that \(H_{F}\) intersects \(B\cap\sigma\) since otherwise \(F\cap B\cap\sigma\) is empty and there is nothing to prove. First, we show that if \(H_{F}\) is within a distance of \(\lambda\) from \(p\), \(H_{F}\cap B\cap\sigma\) is a single open curve. Consider the disk \(H_{F}\cap B\). Let \(\vec{d}\) be the projection of \(\mathbf{n}_{p}\) onto \(H_{F}\). Let \(L\subset H_{F}\) be the line through the center of \(H_{F}\cap B\) orthogonal to \(\vec{d}\). Let \(x\) be any point in \(H_{F}\cap B\cap\sigma\). The angle between \(px\) and the tangent plane at \(p\) is at most \(\arcsin(1/32)\) by Lemma 10(i.b). We already proved that \(\angle(\mathbf{n}_{p},H_{F})\leqslant\arcsin(1/16)\). We are interested in an upper bound on the distance \(d(x,L)\). Let \(x^{*}\) be the orthogonal projection of \(x\) onto \(L\). Consider the triangle \(pxx^{*}\). Observe that \(\angle xpx^{*}\) is the angle between \(px\) and the tangent plane at \(p\) and \(\angle px^{*}x\) is at least \(\pi/2-\angle(\mathbf{n}_{p},H_{F})\). We obtain  
\[d(x,x^{*})\leqslant d(p,x)\frac{\sin\arcsin(1/32)}{\cos\arcsin(1/16)} \leqslant(0.035)(2\lambda)\leqslant 0.07\lambda.\]  
Let \(L^{*}\subset H_{F}\) be the strip of points at distance \(0.07\lambda\) or less from \(L\). Since the radius of \(B\) is \(2\lambda\) and \(H_{F}\) is at most \(\lambda\) distance from \(p\), the radius of \(H_{F}\cap B\) is at least \(\sqrt{3}\lambda\). It follows that the boundary of \(H_{F}\cap B\) intersects \(L^{*}\) in two disjoint circular arcs, say \(A\) and \(A^{\prime}\). We already proved that there is no closed curve in \(H_{F}\cap B\cap\sigma\) (Lemma 11). Now we show that \(H_{F}\cap B\cap\sigma\) cannot contain two open curves. First, we eliminate the case where \(H_{F}\cap B\cap\sigma\) are two open curves each having an end point on \(\operatorname{bd}\sigma\). For this to happen, \(F\) has to intersect \(\operatorname{bd}\sigma\) in more than one point, which is impossible by Lemma 1 if the protecting balls satisfy condition C1-C3. Therefore, we can assume that if \(H_{F}\cap B\cap\sigma\) has two open curves, at least one, say \(C\), has both end points on the circular arcs \(A\) and/or \(A^{\prime}\). Consider the case where \(C\) has end points both in \(A\) and \(A^{\prime}\). The other curve, say \(C^{\prime}\), in \(H_{F}\cap B\cap\sigma\) must have an end point in at least one of \(A\) and \(A^{\prime}\). It follows that either \(A\) or \(A^{\prime}\) should have an end point of both \(C\) and \(C^{\prime}\).