Minor point: can anyone give a quick argument that the angle bisectors actually have an intersection point? (This is slightly non-trivial -- one must decide if angle bisectors are rays or lines first.) --JBL 11:20, 7 September 2006 (EDT)

I think that the section "Proof of Existance" does that. However, it assumes that the locus of points equidistant from the two lines is the angle bisectors, and also that any two given angle bisectors intersect. But it is fairly easy to see that if we assume that two of them are parallel, then we have a degenerate triangle. (We can also say that they intersect at the point of infinity, which is then the circumcenter of the triangle, which is true in the projective plane.) —Boy Soprano II 17:05, 7 September 2006 (EDT)

Sorry, that question was very poorly phrased. What I meant was the proof that the three angle bisectors are concurrent implicitly assumes that any two angle bisectors will intersect. While this is true, it is not as obvious as it is in the case of the perpendicular bisectors, say. In particular, it is complicated by the fact that an angle is defined by 2 rays, and as such the angle bisector is a ray (not a line) and so there are additional ways to make two angle bisectors not intersect than to have them be parallel.

One way out of this is to use that fact that any two interior cevians of a triangle intersect. Another way would be to just treat the angle bisectors as lines instead of rays, but include an explanation of why two of them should intersect. Either way, I think the phrase, "Let I be the intersection of the angle bisectors of angles B and C" should be accompanied by some brief explanation of why I exists. (I probably would leave out the bit about the point at infinity, even though it's nice -- I think I'm making this complicated enough without that.) --JBL 20:48, 7 September 2006 (EDT)

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