van Obel's Theorem in hyperbolic geometry
If the point X does not lie on any side of the hyperbolic ΔABC, and
AB meets CX in P, BC meets AX in Q and CA meets BX in R,
then
h(A,X,Q) = cosh(d(B,Q)h(A,P,B)+cosh(d(Q,C))h(A,R,C),
h(B,X,R) = cosh(d(A,R))h(B,P,A)+cosh(d(R,C))h(B,Q,C),
h(C,X,P) = cosh(d(A,P))h(C,R,A)+cosh(d(P,B))h(C,R,B).
Proof
It is enough to prove the first, the others can be deduced by permuting
the labels.
Although the final result is independent of whether each cut is internal
or external, we have to consider the cases separately to some extent.
The hyperbolic version of Menelaus's Theorem, applied to ΔABQ gives
h(A,P,B)h(B,C,Q)h(Q,X,A) = 1. As h(W,Y,Z)=1/h(Z,Y,W), we get
h(A,P,B) = h(Q,C,B)h(A,X,Q).
Likewise, from ΔAQC, we get h(A,P,C) = h(Q,B,C)h(A,X,Q).
(1) Q lies between B and C.
Then C is not between B and Q, and B not between C and Q.
Thus, h(Q,C,B) = sinh(d(Q,C))/sinh(d(C,B)),
and, h(Q,B,C) = sinh(d(Q,B))/sinh(d(B,C)).
Also, d(B,C) = d(B,Q)+d(Q,C).
Then, as sinh(a)cosh(b)+sinh(b)cosh(a) = sinh(a+b),
cosh(d(B,Q))h(A,P,B)+cosh(d(Q,C))h(A,R,C) = h(A,X,Q)
(2) Q lies beyond C.
Now C lies between B and Q, B does not lie between C and Q.
Thus, h(Q,C,B) = +sinh(d(Q,C))/sinh(d(C,B)),
and, h(Q,B,C) = sinh(d(Q,B))/sinh(d(B,C)).
Also d(B,C)= d(B,Q)d(Q,C).
Then, as sinh(a)cosh(b)sinh(b)cosh(a) = sinh(ab),
cosh(d(B,Q))h(A,P,B)+cosh(d(Q,C))h(A,R,C) = h(A,X,Q)
(3) Q lies beyond B.
This is obtained from case (2) by interchanging {B,R} and {C,P}.

