a theorem on saccheri quadrialterals
Suppose that ABDC is a saccheri quadrialteral in the hyperbolic plane, having
right angles at C and D, and with d(A,C) = d(B,D) = d, d(C,D) = c, d(A,B) = a.
Let h be the length of the altitude of ABDC. Then
(1) sinh(½a) = cosh(d)sinh(½c),
(2) cosh(h)cosh(½a) = cosh(d)cosh(½c).
proof
We may as well move the picture so C,D lie on the real axis, as shown.
We have indicated the hypercircle XY on which A,B lie.
(1) Let x = d(B,C), and let θ be the hyperbolic angle BCD.
Then as ACD is a right angle, cos(ACB) = sin(θ).
We use various results on hyperbolic triangles.
By the Cosine Rule for ΔABC, cosh(a) = cosh(x)cosh(d)sinh(x)sinh(d)sin(θ).
By the Sine Formula for ΔBCD, sin(θ) = sinh(d)/sinh(x).
By Pythagoras Theorem for ΔBCD, cosh(x) = cosh(c)cosh(d).
Combining these, cosh(a) = cosh(c)cosh^{2}(d)sinh^{2}(d).
Using the formulae cosh(t) = 2sinh^{2}(½t)+1, cosh^{2}(d)sinh^{2}(d) = 1, we get
sinh^{2}(½a) = cosh(^{2}(d)sinh^{2}(½c). The result follows.
Let M, N be the midpoints of AB, CD, so MN is the altitude.
Let NB have hyperbolic length y.
By Pythagoras's Theorem applied to NDB, NMB, we get
cosh(y) = cosh(d)cosh(½c) = cosh(h)cosh(½a). The result follows.


arc length on a hypercircle
Suppose that A,B lie on L, a hypercircle of width d associated with the hyperbolic
line H. Let C,D be the feet of the perpendiculars from A,B to H, and let
a=d(A,B), c=d(C,D). Then L, the hyperbolic length of the arc AB of L, satisfies
(1) L = cosh(d)c, and
(2) sinh(½a) = cosh(d)sinh(L/2cosh(d)).
proof
We may as well use the above picture.
Suppose we divide CD into N equal hyperbolic segments. Each has length c/N.
By the above result, the chord above each segment has length l, where
sinh(½l) = cosh(d)sinh(c/2N).
There are N chords, so the sum of their lengths is L*= Nl.
Thus sinh(L*/2N) = cosh(d)sinh(c/2N). Multiplying through by 2N, we get
2Nsinh(L*/2N) = cosh(d).2Nsinh(c/2N).
As N tends to infinity, L* tends to arc length, L and we get L= cosh(d)c.
Finally, from the previous result, sinh(½a) = cosh(d)sinh(½c).
Substituting for c, we get (2).

