Toponogov hinge comparison

Setup and notation. Take the data of the statement. The included angle satisfies $\cos \theta =g_p({\sigma }_1^{\mathrm{\prime }}(0),{\sigma }_2^{\mathrm{\prime }}(0))$ with unit vectors ${\sigma }_1^{\mathrm{\prime }}(0),{\sigma }_2^{\mathrm{\prime }}(0)$, and $c=d_g(x,y)\ge 0$; the triangle inequality gives $\mathrm{\mid }a-b\mathrm{\mid }\le c\le a+b\mathrm{.}$ When $k>0$ put $D:=D_k=\pi \mathrm{/}\sqrt{k}$ and record the hypotheses $a,b,c<D$ and $a+b+c<2D$. Both legs are minimizing, so $a=d_g(p,x)$ and $b=d_g(p,y)$. No geodesic is ever used beyond its stated role: ${\sigma }_1$ and ${\sigma }_2$ are the hinge legs, and ${\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$, ${\sigma }_2$ are the two curves in the first-variation computations below.

The model hinge. Fix $\bar{p}\in M_k^2$ and a unit vector $u\in T_{\bar{p}}{M}_k^2$; since $T_{\bar{p}}{M}_k^2$ is two-dimensional, choose $w\perp u$ with $\mathrm{\mid }w\mathrm{\mid }=1$ and put $v:=(\cos \theta )u+(\sin \theta )w\mathrm{.}$ Then $\mathrm{\mid }v{\mathrm{\mid }}^2={\cos }^2\theta +{\sin }^2\theta =1$ and $g_{\bar{p}}(u,v)=\cos \theta$, so the geodesics $t\mapsto {\exp }_{\bar{p}}(tu)$ and $t\mapsto {\exp }_{\bar{p}}(tv)$ leave $\bar{p}$ at included angle $\theta$. Put $\bar{x}:={\exp }_{\bar{p}}(au),\bar{y}:={\exp }_{\bar{p}}(bv),\widehat{c}:=d_k(\bar{x},\bar{y})\mathrm{.}$ By [F5] the model is complete and every radial geodesic of the model is minimizing on its initial interval up to the cut time; for $k>0$ the cut time is $D$ in every direction at every point, and $a,b<D$, while for $k\le 0$ every radial geodesic minimizes on all of $\mathbb{R}$. Hence $d_k(\bar{p},\bar{x})=a,d_k(\bar{p},\bar{y})=b\mathrm{.}$ The three points $\bar{p},\bar{x},\bar{y}$, together with any minimizing geodesic from $\bar{x}$ to $\bar{y}$ (which exists by F4), therefore form a labelled triple realizing the ordered side lengths $(\widehat{c},b,a)$. The model geodesic used below is the leg ${\bar{\sigma }}_2:={\exp }_{\bar{p}}(\cdot v){\mathrm{\mid }}_{[0,b]}$ from $\bar{p}$ to $\bar{y}$.

The comparison-angle function of the side lengths. Fix $A,B>0$ with $A,B<D$ when $k>0$. Let $I(A,B)$ be the set of $C>0$ for which $(A,B,C)$ satisfies the side hypotheses of [F1]: the strict triangle inequalities and, when $k>0$, $C<D$ and $A+B+C<2D$. We claim $I(A,B)=(\mathrm{\mid }A-B\mathrm{\mid },m(A,B)),m(A,B):=\{\begin{array}{ll}\min \{A+B,2D-A-B\},& k>0,\\ A+B,& k\le 0.\end{array}$ For $k>0$ the conditions $C<D$ and $C<2D-A-B$ combine to $C<\min \{A+B,D,2D-A-B\}$, and since $A,B<D$ one has $\min \{A+B,D,2D-A-B\}=\min \{A+B,2D-A-B\}$: if $A+B\le D$ then $A+B\le D\le 2D-A-B$, and if $A+B\ge D$ then $2D-A-B\le D\le A+B$. For $k\le 0$ there is no upper bound. So $I(A,B)$ is the displayed interval. On it define ${\mathrm{\Phi }}_{A,B}(C)\in (0,\pi )$ by the model cosine law of [F1], i.e. ${\mathrm{\Phi }}_{A,B}(C):=\arccos F_{A,B}(C)$ with $F_{A,B}(C):=\{\begin{array}{ll}{\displaystyle \frac{\cos (\sqrt{k}C)-\cos (\sqrt{k}A)\cos (\sqrt{k}B)}{\sin (\sqrt{k}A)\sin (\sqrt{k}B)}},& k>0,\\ {\displaystyle \frac{A^2+B^2-C^2}{2AB}},& k=0,\\ {\displaystyle \frac{\cosh (\sqrt{-k}A)\cosh (\sqrt{-k}B)-\cosh (\sqrt{-k}C)}{\sinh (\sqrt{-k}A)\sinh (\sqrt{-k}B)}},& k<0.\end{array}$ Each denominator is strictly positive on the domain ($\sin >0$ and $\sinh >0$ for positive arguments, $2AB>0$), so by [F6] the function $F_{A,B}$ is continuous on $I(A,B)$, and ${\mathrm{\Phi }}_{A,B}=\arccos \circ F_{A,B}$ is continuous with values in $(0,\pi )$. Moreover the same function is strictly increasing on $I(A,B)$. On that interval the numerator of $F_{A,B}$ is strictly decreasing in $C$: for $k>0$, $\cos (\sqrt{k}C)$ is strictly decreasing because $\cos$ is strictly decreasing on $[0,\pi ]$ and $C\mapsto \sqrt{k}C$ is strictly increasing with values in $(0,\pi )$; for $k=0$, $A^2+B^2-C^2$ is strictly decreasing in $C>0$; for $k<0$, $-\cosh (\sqrt{-k}C)$ is strictly decreasing because $\cosh$ is strictly increasing on $[0,\mathrm{\infty })$. Hence $F_{A,B}$ is strictly decreasing, and since $\arccos$ is strictly decreasing on $[-1,1]$, the composition ${\mathrm{\Phi }}_{A,B}$ is strictly increasing on $I(A,B)$; in particular ${\mathrm{\Phi }}_{A,B}$ is injective. Moreover ${\mathrm{\Phi }}_{A,B}$ has the one-sided limits ${\lim }_{C\downarrow \mathrm{\mid }A-B\mathrm{\mid }}{\mathrm{\Phi }}_{A,B}(C)=0,{\lim }_{C\uparrow m(A,B)}{\mathrm{\Phi }}_{A,B}(C)=\pi \mathrm{.}$ Indeed at $C=\mathrm{\mid }A-B\mathrm{\mid }$ the addition formulas of [F6] give $F_{A,B}(C)=1$; at $C=A+B$ they give $F_{A,B}(C)=-1$; and when $k>0$ and $m(A,B)=2D-A-B<A+B$, the shift formula $\cos (2\pi -x)=\cos x$ of [F6] gives $\cos (\sqrt{k}C)=\cos (\sqrt{k}(2\pi \mathrm{/}\sqrt{k}-(A+B)))=\cos (\sqrt{k}(A+B))$, so again $F_{A,B}(C)=-1$. Continuity of $F_{A,B}$ and of $\arccos$ gives the two displayed limits, and the values lie in $(0,\pi )$ for $C\in I(A,B)$.

The inverse of $\mathrm{\Phi }$ is the model opposite side. We show that for every $\theta \in (0,\pi )$ the number $\widehat{c}$ of step 1.2 satisfies $\widehat{c}={\mathrm{\Phi }}_{a,b}^{-1}(\theta ),$ which proves at once that $\widehat{c}$ depends only on $(a,b,\theta )$ and that $c_k(a,b,\theta )={\mathrm{\Phi }}_{a,b}^{-1}(\theta )$ for $\theta \in (0,\pi )$. First, $\widehat{c}\le m(a,b)$: the triangle inequality in the model gives $\widehat{c}\le a+b$; when $k>0$ and $a+b>D$ (the only case in which $2D-a-b<a+b$) let ${\bar{p}}^{\ast }$ be the antipode of $\bar{p}$. By [F5] the cut time of $\bar{p}$ is $D$ in every direction with cut locus $\{{\bar{p}}^{\ast }\}$, so the radial geodesic $\phi (t):={\exp }_{\bar{p}}(tu)$ runs from $\bar{p}$ to ${\bar{p}}^{\ast }$ and is minimizing on $[0,D]$, and it is $2D$-periodic: $\phi (s+D)$ is the cut point of $\phi (s)$ in direction $u$, hence the antipode of $\phi (s)$, and applying this twice with the involution $q\mapsto -q$ gives $\phi (s+2D)=\phi (s)$. Consequently for $0\le s<t\le s+D$ one has $d_k(\phi (s),\phi (t))=t-s$, because the radial geodesic from $\phi (s)$ in direction $u$ is minimizing on all of $[0,D]$. Since $\bar{x}=\phi (a)$ and $t\mapsto {\exp }_{\bar{p}}(tv)$ is minimizing on $[0,D]$ with terminal point ${\exp }_{\bar{p}}(Dv)={\bar{p}}^{\ast }$, the cut point at time $D$, the sub-segments of $\phi$ from $a$ to $D$ and of this radial geodesic from $b$ to $D$ give $d_k(\bar{x},{\bar{p}}^{\ast })=D-a$ and $d_k(\bar{y},{\bar{p}}^{\ast })=D-b$, whence $\widehat{c}\le (D-a)+(D-b)=2D-a-b$. So $\widehat{c}\le m(a,b)$ in all cases. Second, $\widehat{c}>\mathrm{\mid }a-b\mathrm{\mid }$ and $\widehat{c}<m(a,b)$ unless the included angle is $0$ or $\pi$. If $\widehat{c}=a+b$, the concatenation of the two model legs $\bar{x}\bar{p}$ and $\bar{p}\bar{y}$ is a piecewise smooth curve of length $a+b=d_k(\bar{x},\bar{y})$, hence minimizing; by F4 it is a unit-speed geodesic, so its one-sided velocities at $\bar{p}$ agree. The velocity arriving from the $\bar{x}$-side is $-u$ and the departing velocity is $v$, so $v=-u$ and $\cos \theta =-1$, i.e. $\theta =\pi$. Similarly, if $\widehat{c}=\mathrm{\mid }a-b\mathrm{\mid }$, say $a\ge b$, then $d_k(\bar{p},\bar{x})=b+d_k(\bar{y},\bar{x})$, so a minimizing geodesic from $\bar{p}$ to $\bar{y}$ concatenated with a minimizing geodesic from $\bar{y}$ to $\bar{x}$ is minimizing and smooth; since $d_k(\bar{p},\bar{y})=b<D$ when $k>0$ and minimizing geodesics of the model are unique by [F5] (and for $k\le 0$ likewise), that initial segment is the leg $t\mapsto {\exp }_{\bar{p}}(tv)$, while as a segment of the geodesic from $\bar{p}$ to $\bar{x}$ it is also $t\mapsto {\exp }_{\bar{p}}(tu)$; hence $v=u$ and $\theta =0$. Finally, if $\widehat{c}=2D-a-b$ with $k>0$, the same concatenation argument applied to the two minimizing pieces $\bar{x}\to {\bar{p}}^{\ast }$ and ${\bar{p}}^{\ast }\to \bar{y}$ shows that these pieces join smoothly at ${\bar{p}}^{\ast }$; the first is the sub-segment $\phi {\mathrm{\mid }}_{[a,D]}$ of the geodesic $\phi$, uniqueness of minimizing geodesics between those two points ([F5]; the distance is $D-a<D$) identifies the pieces with the corresponding sub-segments of $\phi$, and smoothness forces $\bar{y}=\phi (2D-b)=\phi (-b)={\exp }_{\bar{p}}(-bu)$. Since also $\bar{y}={\exp }_{\bar{p}}(bv)$ with $b<D$, the uniqueness of minimizing geodesics in the model gives $v=-u$, so $\theta =\pi$. Thus for $\theta \in (0,\pi )$ we have $\mathrm{\mid }a-b\mathrm{\mid }<\widehat{c}<m(a,b)$, i.e. $\widehat{c}\in I(a,b)$. Now the triple $(\bar{p},\bar{x},\bar{y})$ realizes the ordered side lengths $(\widehat{c},b,a)$ with minimizing sides: the two legs are minimizing geodesics of lengths $a$ and $b$, and a minimizing geodesic from $\bar{x}$ to $\bar{y}$ exists by F4. Hence it is a comparison triangle with these ordered side lengths, and by [F1] its angle at $\bar{p}$ is the model cosine law value ${\mathrm{\Phi }}_{a,b}(\widehat{c})$; its angle at $\bar{p}$ is $\theta$ by construction of $v$. Therefore $\theta ={\mathrm{\Phi }}_{a,b}(\widehat{c})$, and injectivity of ${\mathrm{\Phi }}_{a,b}$ from step 1.3 gives $\widehat{c}={\mathrm{\Phi }}_{a,b}^{-1}(\theta )$. In particular the model opposite side $c_k(a,b,\theta )$ is well defined on $(0,\pi )$ and equals ${\mathrm{\Phi }}_{a,b}^{-1}(\theta )$.

The endpoint values. At $\theta =0$ the construction of step 1.2 gives $v=u$, so $\bar{x}=\phi (a)$, $\bar{y}=\phi (b)$ for the radial geodesic $\phi (t)={\exp }_{\bar{p}}(tu)$, which minimizes on $[0,\max \{a,b\}]$ by [F5]; therefore $c_k(a,b,0)=d_k(\phi (a),\phi (b))=\mathrm{\mid }a-b\mathrm{\mid }\mathrm{.}$ At $\theta =\pi$ one has $v=-u$, so $\bar{y}=\phi (-b)$, where $\phi$ minimizes on all of $\mathbb{R}$ when $k\le 0$ and on $[0,D]$ together with its $2D$-periodicity when $k>0$. If $k\le 0$, then $c_k(a,b,\pi )=d_k(\phi (a),\phi (-b))=a+b$. If $k>0$ and $a+b\le D$, the sub-interval $[-b,a]$ has length $a+b\le D$, so it is minimizing and $c_k(a,b,\pi )=a+b$; if $a+b>D$, periodicity gives $\bar{y}=\phi (2D-b)$ with $(2D-b)-a=2D-a-b\in (0,D)$, so $c_k(a,b,\pi )=2D-a-b$. Thus $c_k(a,b,\pi )=\{\begin{array}{ll}\min \{a+b,2D-a-b\},& k>0,\\ a+b,& k\le 0.\end{array}$ This also shows that $c_k(a,b,\theta )$ is well defined on all of $[0,\pi ]$, agrees with the model-hinge construction of the statement, and extends ${\mathrm{\Phi }}_{a,b}^{-1}$ at the two endpoints.

The degenerate cases $c=a+b$ and $c=\mathrm{\mid }a-b\mathrm{\mid }$. By step 1.1 exactly one of the following holds: (i) $c=a+b$; (ii) $c=\mathrm{\mid }a-b\mathrm{\mid }$; (iii) $\mathrm{\mid }a-b\mathrm{\mid }<c<a+b$. We dispose of the first two cases. If $c=a+b$, the concatenation of ${\sigma }_1$ reversed and ${\sigma }_2$ is a piecewise smooth curve from $x$ to $y$ of length $a+b=d_g(x,y)$, hence minimizing; by F4 it is a unit-speed geodesic, so its one-sided velocities at $p$ agree: the velocity arriving from the $x$-side is $-{\sigma }_1^{\mathrm{\prime }}(0)$ and the departing velocity is ${\sigma }_2^{\mathrm{\prime }}(0)$, so ${\sigma }_2^{\mathrm{\prime }}(0)=-{\sigma }_1^{\mathrm{\prime }}(0)$ and $\theta =\pi$. The endpoint value of step 2.2 then gives $c_k(a,b,\theta )=c_k(a,b,\pi )=a+b=c$ when $k\le 0$; when $k>0$ the hypothesis gives $2(a+b)=a+b+c<2D$, i.e. $a+b<D$, so $2D-a-b>a+b$ and $c_k(a,b,\pi )=\min \{a+b,2D-a-b\}=a+b=c$. So $c\le c_k(a,b,\theta )$ in case (i). If $c=\mathrm{\mid }a-b\mathrm{\mid }$, the reverse triangle inequality in the model, applied to the hinge points $\bar{p},\bar{x},\bar{y}$ of step 1.2, gives $c=\mathrm{\mid }a-b\mathrm{\mid }=\mid d_k(\bar{p},\bar{x})-d_k(\bar{p},\bar{y})\mid \le d_k(\bar{x},\bar{y})=c_k(a,b,\theta ),$ which is the claim. Only case (iii) remains.

Case (iii): the strict comparison data. Assume $\mathrm{\mid }a-b\mathrm{\mid }<c<a+b$. Then the triple $(a,b,c)$ satisfies the strict triangle inequalities, and for $k>0$ the bounds $a,b,c<D$ and $a+b+c<2D$ are hypotheses; hence $(a,b,c)$ satisfies the side hypotheses of [F1] and the comparison angle $\bar{\theta }:={\mathrm{\Phi }}_{a,b}(c)\in (0,\pi )$ is defined, with $c={\mathrm{\Phi }}_{a,b}^{-1}(\bar{\theta })$ by step 1.3. Also $\theta \mathrm{\ne }0$: if $\theta =0$ then ${\sigma }_1^{\mathrm{\prime }}(0)={\sigma }_2^{\mathrm{\prime }}(0)$ by [F6] (equality in the Cauchy–Schwarz bound for unit vectors), so the two legs coincide by F4, placing $x$ and $y$ on one radial geodesic and forcing $c=\mathrm{\mid }a-b\mathrm{\mid }$, contrary to case (iii). If $\theta =\pi$, then $c<a+b$ by case (iii) and $c<2D-a-b$ when $k>0$ by hypothesis, so $c<\min \{a+b,2D-a-b\}=c_k(a,b,\pi )=c_k(a,b,\theta )$ by step 2.2 when $k>0$, and $c<a+b=c_k(a,b,\pi )=c_k(a,b,\theta )$ when $k\le 0$; this is the claim. It remains to treat $\theta \in (0,\pi )$, which is done in the next layer.

The shifted configuration. Assume case (iii) and $\theta \in (0,\pi )$. For $\epsilon \in (0,a)$ put $x_{\epsilon }:={\sigma }_1(a-\epsilon ),c_{\epsilon }:=d_g(x_{\epsilon },y)\mathrm{.}$ Then $d_g(p,x_{\epsilon })=a-\epsilon$, the sub-segment ${\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$ is a minimizing geodesic from $p$ to $x_{\epsilon }$, and $d_g(x_{\epsilon },x)=\epsilon$. By the triangle inequality, $c_{\epsilon }\to c$ and $c-\epsilon \le c_{\epsilon }\le c+\epsilon$. For all sufficiently small $\epsilon >0$ the triple $(a-\epsilon ,b,c_{\epsilon })$ satisfies the side hypotheses of [F1]: $c_{\epsilon }>0$ and the strict triangle inequalities hold by the estimates $c_{\epsilon }\ge c-\epsilon$, $c_{\epsilon }\le c+\epsilon$ together with $\mathrm{\mid }a-b\mathrm{\mid }<c<a+b$ (for the inequality $c_{\epsilon }<(a-\epsilon )+b$ use $c_{\epsilon }\le c+\epsilon <a+b-\epsilon$ for $\epsilon <\frac{1}{2}(a+b-c)$; the lower bounds are analogous), and for $k>0$ one has $a-\epsilon ,b,c_{\epsilon }<D$ and $(a-\epsilon )+b+c_{\epsilon }\le a+b+c<2D$. Also $p\mathrm{\notin }\mathrm{Cut}(x_{\epsilon })$: if $p\in \mathrm{Cut}(x_{\epsilon })$, then by the characterization F4 either $x_{\epsilon }$ and $p$ are conjugate along a minimizing geodesic joining them, or two distinct minimizing geodesics join them. In the second case, reversing the two geodesics gives two distinct minimizing geodesics joining $p$ and $x_{\epsilon }$; at most one of these two is the segment ${\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$, so choose one of them and call it $\tau$. Then $\tau :[0,a-\epsilon ]\to M$ is a unit-speed minimizing geodesic with $\tau (0)=p$ and $\tau (a-\epsilon )=x_{\epsilon }$, and ${\tau }^{\mathrm{\prime }}(0)\mathrm{\ne }{\sigma }_1^{\mathrm{\prime }}(0)$, because otherwise $\tau ={\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$ by uniqueness of geodesics with the same initial data F4, contrary to the choice of $\tau$. Since ${\sigma }_1(t)={\exp }_p(t{\sigma }_1^{\mathrm{\prime }}(0))$, applying the converse (b)(2) of the characterization F4 with base $p$, direction ${\sigma }_1^{\mathrm{\prime }}(0)$ and time $a-\epsilon$ gives $c_p({\sigma }_1^{\mathrm{\prime }}(0))\le a-\epsilon$, contradicting $c_p({\sigma }_1^{\mathrm{\prime }}(0))\ge a$, which holds because ${\sigma }_1{\mathrm{\mid }}_{[0,a]}$ is minimizing. In the first case, if the conjugacy is carried by a minimizing geodesic joining $p$ and $x_{\epsilon }$ other than ${\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$, then that geodesic together with ${\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$ gives two distinct minimizing geodesics joining $p$ and $x_{\epsilon }$, and we are back in the second case. If instead the conjugacy is carried by ${\sigma }_1$, then $p$ and ${\sigma }_1(a-\epsilon )$ are conjugate along ${\sigma }_1$ with $a-\epsilon <a$, so ${\sigma }_1{\mathrm{\mid }}_{[0,a]}$ does not minimize past its first conjugate point, by A geodesic does not minimize past its first conjugate point; this contradicts the minimality of ${\sigma }_1$. Hence $p\mathrm{\notin }\mathrm{Cut}(x_{\epsilon })$.

The support inequality for the shifted triangle. Take a minimizing geodesic ${\gamma }_{\epsilon }$ from $x_{\epsilon }$ to $y$ (exists by F4). Apply [F2] with $N:=M$, $\gamma :={\sigma }_2$ (a unit-speed minimizing geodesic of length $b$ from $p$ to $y$), $o:=x_{\epsilon }$, $a:=a-\epsilon =d_g(x_{\epsilon },p)$ and $b:=c_{\epsilon }=d_g(x_{\epsilon },y)$: by step 5.1 the triples $(b,c_{\epsilon },a-\epsilon )$ are the side lengths of a comparison triangle $({\bar{x}}_{\epsilon },\bar{p},\bar{y})$ in $M_k^2$, with model side ${\bar{\sigma }}_{2,\epsilon }$ from $\bar{p}$ to $\bar{y}$. The lemma gives $d_g(x_{\epsilon },{\sigma }_2(t))\ge d_k({\bar{x}}_{\epsilon },{\bar{\sigma }}_{2,\epsilon }(t))(0\le t\le b)\mathrm{.}$

The two first-variation derivatives. Apply [F3] in $M$ with $o:=x_{\epsilon }$, $q:=p$ (so the role of the minimizing geodesic $\sigma$ of [F3] is played by the reverse of ${\sigma }_1{\mathrm{\mid }}_{[0,a-\epsilon ]}$) and the moving geodesic $\gamma :={\sigma }_2$. The hypothesis holds by step 5.1, and the included angle at $p$ is $\theta$, because the reversed sub-segment has the same initial direction at $p$ as ${\sigma }_1$. Hence ${\frac{d}{dt}\mid }_{0^{+}}{d}_g(x_{\epsilon },{\sigma }_2(t))=-\cos \theta \mathrm{.}$ Apply [F3] in the model $M_k^2$ with $o:={\bar{x}}_{\epsilon }$, $q:=\bar{p}$ (the minimizing geodesic from ${\bar{x}}_{\epsilon }$ to $\bar{p}$ being the corresponding side of the comparison triangle) and $\gamma :={\bar{\sigma }}_{2,\epsilon }$. The vertex $\bar{p}$ is not a cut point of ${\bar{x}}_{\epsilon }$: when $k>0$ the model distance is $d_k({\bar{x}}_{\epsilon },\bar{p})=a-\epsilon <D$ while the cut locus of a point of the model sphere is the singleton antipode at distance $D$ [F5]; when $k\le 0$ the model has empty cut loci [F5]. Hence ${\frac{d}{dt}\mid }_{0^{+}}{d}_k({\bar{x}}_{\epsilon },{\bar{\sigma }}_{2,\epsilon }(t))=-\cos {\bar{\theta }}_{\epsilon },{\bar{\theta }}_{\epsilon }:={\mathrm{\Phi }}_{a-\epsilon ,b}(c_{\epsilon }),$ the equality ${\bar{\theta }}_{\epsilon }={\mathrm{\Phi }}_{a-\epsilon ,b}(c_{\epsilon })$ holding because the angle at $\bar{p}$ of the comparison triangle is its comparison angle by [F1], and the two legs at $\bar{p}$ are the side of length $a-\epsilon$ and the side ${\bar{\sigma }}_{2,\epsilon }$ of length $b$.

The angle comparison. Define $G_{\epsilon }(t):=d_g(x_{\epsilon },{\sigma }_2(t))-d_k({\bar{x}}_{\epsilon },{\bar{\sigma }}_{2,\epsilon }(t))$ on $[0,b]$. By step 6.1, $G_{\epsilon }\ge 0$, and $G_{\epsilon }(0)=(a-\epsilon )-(a-\epsilon )=0$. Both summands have right derivatives at $0$ by step 6.2, so $G_{\epsilon }$ has a right derivative at $0$, and since $G_{\epsilon }(t)\ge 0=G_{\epsilon }(0)$ for $t>0$, $0\le G_{\epsilon }^{\mathrm{\prime }}(0^{+})=\cos {\bar{\theta }}_{\epsilon }-\cos \theta \mathrm{.}$ Since $\cos$ is strictly decreasing on $[0,\pi ]$ and ${\bar{\theta }}_{\epsilon }\in (0,\pi )$, $\theta \in (0,\pi )$, this gives $\theta \ge {\bar{\theta }}_{\epsilon }$.

Passage to the limit and conclusion in case (iii). As $\epsilon \downarrow 0$ we have $a-\epsilon \to a$ and $c_{\epsilon }\to c$ by step 5.1. The three explicit formulas of step 1.3 define a function of $(A,B,C)$ that is jointly continuous near $(a,b,c)$ by [F6]: the denominators $\sin (\sqrt{k}A)\sin (\sqrt{k}B)$, $2AB$ and $\sinh (\sqrt{-k}A)\sinh (\sqrt{-k}B)$ are nonzero there, and for $k>0$ one has $\sqrt{k}a,\sqrt{k}b\in (0,\pi )$ because $a,b<D$; also $\arccos$ is continuous. Hence ${\bar{\theta }}_{\epsilon }={\mathrm{\Phi }}_{a-\epsilon ,b}(c_{\epsilon })⟶{\mathrm{\Phi }}_{a,b}(c)=\bar{\theta }\mathrm{.}$ Step 7.1 therefore gives $\theta \ge \bar{\theta }$. Since $c={\mathrm{\Phi }}_{a,b}^{-1}(\bar{\theta })$ (step 4.1), $\theta \in (0,\pi )$ and ${\mathrm{\Phi }}_{a,b}^{-1}$ is strictly increasing (step 1.3), while $c_k(a,b,\theta )={\mathrm{\Phi }}_{a,b}^{-1}(\theta )$ by step 2.1, $c={\mathrm{\Phi }}_{a,b}^{-1}(\bar{\theta })\le {\mathrm{\Phi }}_{a,b}^{-1}(\theta )=c_k(a,b,\theta )\mathrm{.}$ This completes case (iii) and, with the degenerate cases of step 3.1, the whole proof.