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We all know that all three altitudes of a triangle meets in the orthocenter of the triangle. It's a quite classical problem and is proven. However, what I really wanna know is what characteristic of the triangle is the profound for this to happen? E.g: Is this because of the sum of 3 internal angles equals 180? In Non-Euclidean geometry, where sum of 3 internal angles is greater or smaller than 180 degree, does the 3 altitudes meets in a single point? Or is it because of another reason?

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Euler has quite a few things named after him, so there is no need to deprive Euclid of any honors! :D –  Mariano Suárez-Alvarez Feb 19 '13 at 7:43
In the hyperbolic plane, the perpendicular bisectors of the sides of a triangle can be concurrent, have a common perpendicular or be asymptotic. In general, then, there is no orthocenter. –  Mariano Suárez-Alvarez Feb 19 '13 at 7:53
+1 Damn intresting observation! –  Arjang Feb 19 '13 at 7:54
Have you heard, that the orthocenter is isogonal conjugate of circumcentre? It does not explain everything about the orthocenter, but for me it does explain a lot (isogonal conjugation contains traces of $z \mapsto z^{-1}$ transformation). –  dtldarek Feb 19 '13 at 8:11
Something else that bothers me more is that the existence of the Euler line: the orthocenter, centroid, and circumcenter of any triangle are always collinear. It sounds interesting but doesn't it seem a little bit coincident? I mean I know how to prove it but it doesn't sound natural to me. –  Nhím Hổ Báo Feb 19 '13 at 15:37
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It is perhaps interesting to note that the definition of altitudes is perfectly straightforward for simplexes in higher dimensions, but that already in dimension $3$ the altitudes of a general tetrahedron are not concurrent. For instance for the tetrahedron with vertices $(0,0,0)$, $(1,0,0)$, $(0,1,0)$, $(1,0,1)$, two of the altitudes meet in the origin and two others meet in $(1,0,0)$, but there are no other points of intersection; in general position none of the altitudes will intersect.

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The proofs I know all use Euklidean geometry (e.g. the orthocenter is the intersection of the middle orthogonals for a bigger triangle).

In synthetic geometry, one can consider translation planes with an orthogonality relation and the Fano axiom (diagonals of a nondegenerate parallelogram intersect), thus minimally allowing the proof above. One can show that this makes the geometry at least a Pappus plane.

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What is synthetic geometry? –  Arjang Feb 20 '13 at 2:57
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Vladimir Arnol'd often emphasized that it was because of the Jacobi identity in Lie algebras. I think he might have meant the algebra so(3) represented as $R^3$ with vector cross-product as the multiplication, but am not sure, and it would be nice to see his remark explained.

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