proof of theorem HC2

 Theorem HC2 If L and M are the hyperbolic circles K(C,r) and K(D,s), and d(C,D) = t, then the hyperbolic radical axis hR(L ,M) is (1) a hyperbolic line perpendicular to CD if e-t < cosh(r)/cosh(s) < et, and (2) the empty set otherwise. A key step requires a lemma whose proof consists of a series of calculations. It is probably best to omit the proof at first reading. Lemma Suppose that A, B and Q lie on a hyperbolic line H with d(A,B) = t > 0. Then (1) e-t < cosh(d(A,Q))/cosh(d(B,Q) < et, and (2) for s ε (e-t,et), there is a unique Q on H with cosh(d(A,Q))/cosh(d(B,Q)) = s. Proof of Theorem The proof consists of three steps: (a) If Q ε hR(L,M), and Q* is the foot of the hyperbolic perpendicular from Q to the hyperbolic line CD, then R ε hR(L,M). (b) There is at most one point R in hR(L,M) which lies on CD. (c) If the point R as in (b) exists, then the perpendicular to CD through R is the hR(L,M). (a) Suppose that Q* is the foot of the hyperbolic perpendicular from a point Q to the hyperbolic line CD. Let d(C,Q*) = q*, d(C,Q) = q, d(Q,Q*) = h. Then hp(Q,L) = cosh(q)/cosh(r), hp(Q*,L) = cosh(q*)/cosh(r). By the hyperbolic version of Pythagoras's Theorem cosh(q) = cosh(q*)cosh(h). Then hp(Q,L) = hp(Q*,L)cosh(h). Similarly, hp(Q,M) = hp(Q*,M)cosh(h). Thus, if Q ε hR(L,M) i.e. hp(Q,L) = hp(Q,M), then hp(Q*,L) = hp(Q*,M), so that Q* ε hR(L,M). (b) Now, R ε hR(L,M) lies on CD, if and only if hp(R,L) = hp(R,M). The condition amounts to cosh(d(R,C))/cosh(r) = cosh(d(R,D))/cosh(s). i.e. cosh(d(R,C))/cosh(d(R,D)) = cosh(r)/cosh(s). By the Lemma, this is possible only if e-t < cosh(r)/cosh(s) < et, where t = d(C,D). When the ineqalities are satifsied, R is unique. (c) If we have a point R as in (b), then reversing the argument in (a), we see that each point of the hyperbolic line through R perpendicular to CD lies in hR(L,M). By (a), each point on hR(L,M) projects to R, so lies on this hyperbolic perpendicular.