Proof of the Altitudes Theorem for Hyperbolic Triangles

An h-altitude of a hyperbolic triangle is an h-segment
through a vertex perpendicular to the opposite side.

The Altitudes Theorem for Hyperbolic Triangles
If any two h-altitudes of a hyperbolic triangle meet,
then the h-altitudes are concurrent.

Let ABC be an h-triangle, and let the h-altitudes be
AQ, BR and CP.

Since the labelling of the vertices is immaterial, we may as
well assume that <ABC is the largest angle of the h-triangle.

If <ABC is a right angle, then AB and CB are h-altitudes,
so all three h-altitudes meet at B.

In the CabriJava figure on the right, we can move B to see
that, if <ABC is acute, then P and Q lie on the sides.
while if <ABC is obtuse, then they lie outside the h-triangle.

This is an easy consequence of the very important
Hyperbolic Triangle Theorem.

Thus, we see that, either all the h-ratios are positive, or just
two are negative, i.e. their product is always positive.

Consider the case where the angles are acute.
By the Hyperbolic Tangent Formula, since <BQA = π/2,
sinh(d(B,Q)) = tanh(d(A,Q))/tan(<ABC), and
sinh(d(Q,C)) = tanh(d(A,Q))/tan(<ACB), so that
h(B,Q,C) = tan(<ACB)/tan(<ABC).
h(A,P,B) = tan(<CBA)/tan(<CAB), and
h(C,R,A) = tan(<BAC)/tan(<BCA).
It follows that the product is 1 in this case.

If the angle at B is obtuse, then we need to replace
this angle by its complement in the Tangent Formula.
Once again, we see that the product of h-ratios is 1.

Thus, by the Converse of Ceva's Theorem,
provided two of the h-altitudes meet,
the h-altitudes AQ, BR and CP are concurrent.

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