van Obel's's Theorem in euclidean geometry

 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(a-b), 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}.