the symmetry lemma

 symmetry lemma Suppose that K is a hyperbolic circle, horocycle or hypercircle, and that A, B lie on K. Let h denote inversion in H, the hyperbolic perpendicular bisector of AB. Then (1) H meets K, (2) h(K) = K, (3) if K is a horocycle at X, then h(X) = X, and (4) if K is a hyperbolic circle with centre C, then h(C) = C. proof By applying the origin lemma, we can assume that the hyperbolic mid-point of AB is at O. Then the hyperbolic segment AB is also the euclidean segment AB, and the hyperbolic bisector is the diameter H perpendicular to AB. The circle defining K passes through A and B, so its euclidean centre lies on the line defining H. It follows that H cuts K. This proves (1). Now return to the general position. Suppose that H cuts K at P. Since H is the hyperbolic bisector of AB, and P is on H, h maps A,B,P to B,A,P, and hence maps the euclidean arc K to itself. This proves (2). (3) is straight-forward since X is the unique point where the circle defining K meets the boundary C, and h fixes K and C. (4) is simply the observation that, if inversion in H maps a hyperbolic circle to itself, then the centre must lie on H.