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.

Suppose that A, B and Q lie on a hyperbolic line H with d(A,B) = t > 0.
(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 lemma

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.

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