inradius theorem
The inradius R of the hyperbolic ΔABC satisfies
(1) tanh(R) = sinh(sa)tan(½α) = sinh(sb)tan(½β) = sinh(sc)tan(½γ), and
(2) tanh(R) = Δ/2sinh(s),
(3) tanh(R) = Φ/4cos(½α)cos(½β)cos(½γ).
proof
We use the diagram on the right, even though it (appears) euclidean.
ΔA'BX has a right angle at A', <A'BX = ½β, d(X,A') = R, d(B,A') = sb,
the last by the lemma. Applying the Tangent Formula,
tan(½β) = tanh(R)/sinh(sb). This gives one part of (1).
The others are similar.
If we apply each part of (1) and multiply by sinh^{3}(s), we get
sinh^{3}(s)tanh^{3}(R) = sinh(sa)tan(½α)sinh(sb)tan(½β)sinh(sc)tan(½γ)sinh^{3}(s).
If we apply (Δ1),(R1), the stheorem and standard trigonometry, we get
sinh^{3}(s)tanh^{3}(R) = Δ^{3}/8. (2) follows at once.
(3) follows from (2), the stheorem and (R1).

