Converse of Ceva's Theorem for Hyperbolic Triangles
If P lies on the h-line AB, Q on BC and R on CA such that
h(A,P,B)h(B,Q,C)h(C,R,A) = 1,
and two of the h-lines CP, BR and AQ meet,
then all three are concurrent.
A little consideration shows that, if we change the labels of the vertices,
then either the factors are simply permuted, or are inverted and permuted.
Since our hypothesis is that the product is 1, the labelling of the vertices
of the h-triangle is immaterial.
The key is to show that the h-line through the intersection X of two of
If P lies between A and B, then CP cuts the h-segment AQ, at X say.
A similar argument applies if R lies between A and C.
Now suppose that P lies beyond B, and R beyond C.
Next suppose that P lies beyond B, and R beyond A.
The case where P is beyond A, and R beyond C is similar.
Note that, in all of the above cases, the h-lines AQ,BR,CP must meet.
We are left with cases where both P and R lie beyond A.
Suppose first that BR and CP meet, at X, say.
Suppose next that BR and AQ meet, at X, say.
The case where CP and AQ meet is similar to the previous case.
Note. There are cases where the h-lines are parallel or ultraparallel.
Note. Essentially the same argument works in the euclidean case.
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