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
midpoint 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 straightforward 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.
