TITLE: Altitudes of a triangle
QUESTION [13 upvotes]: The three altitudes of a triangle are concurrent -- this is true in all three constant curvature geometries (Euclidean, hyperbolic, spherical), but, as far as I know, the proofs are different in the three cases. Is there a "uniform" proof?

REPLY [5 votes]: I think this (and other similar facts) can be derived uniformly using elementary analyticity arguments. First you prove it for the round unit sphere. by rescaling this  implies that it's true for the round sphere of any radius. now look at the cosine law in the simply connected space form of constant curvature $k$. since this formula is analytic in $k$, the "size" of the potential failure of the altitudes to intersect at the same point (measured in any reasonable way) will also be analytic in $k$ and since it's constantly zero for $k>0$ it must be constantly zero for all $k$.
It should not be hard to make the above into a rigorous argument.