证明
Lemma 28 ([Exceptional-incidence identity]). The vectors M_C,x_D lie in a…Lemma 27 ([Fan identities]). For i∈A,B,C,D there is a unit vector M_i∈T such…观点
Take either full line H at D . The path Y_D,H,X_D,H_0 uses two fields of F_D and the loose field, so it is one of the admissible three-step paths. Thus d(Y_D,H_0)≥√3=d(Y_D,c)+d(c,H_0). Equality follows from the triangle inequality. The segment from Y_D to H_0 passes through c , and the directions of y_D,h_0 are antipodal. Let Σ be the completed space of directions of Z at c , and use a hat for the direction of a nonzero vector. Define a=(1/√3), b=(1/3). Both lie strictly between zero and π/2 , and 2a+b=π . Equation gives d_Σ(M_C,h_0)=a, d_Σ(h_0,x_D)=b. The two arcs occur consecutively in the Euclidean sector of Lemma 28 ([Exceptional-incidence identity]); their concatenation is its geodesic direction arc. By Equation, d_Σ(x_D,y_D)=(-1/3)=π-b. The direction of M_D=x_D⊕y_D bisects this last arc, because its two summands have equal length. Consequently d_Σ(x_D,M_D)=a . The concatenation h_0,x_D,y_D has length π and antipodal endpoints. It is therefore minimizing, so its two arcs continue geodesically through x_D . It follows that M_C, h_0, x_D, M_D is a local geodesic in Σ : continuation at h_0 holds inside the first sector, and continuation at x_D holds inside the minimizing semicircle just described. Its length is a+b+a=π . The space Σ is (1) , so makes this local geodesic globally minimizing, including the endpoint case of length exactly π . Equivalently, apply the short local-geodesic assertion to every strict initial subarc and pass to the endpoint by continuity of distance. Its endpoints are antipodal. Since M_C,M_D have norm one, the cone distance and scalar product have the stated values.