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Oct 7, 2021 at 9:30 history made wiki Post Made Community Wiki by Stefan Kohl
Sep 8, 2021 at 9:19 comment added bathalf15320 Of course, the validity of my claim depends on what you mean by a triangle. I thought that I had covered that point in my answer. However, it is a fact that in most of the literature, the so-called degenerate cases are implicitly or explicitly included in many arguments (example: Conway's discussion of whether there can be a direct proof of Steiner-Lehmus).
Sep 7, 2021 at 17:57 comment added Robin Saunders If, for example, we take a definition with "directions of edges" or similar built in, then setting two vertices equal gives a family of triangles with the same vertices but different angles; these can be seen as limits of nondegenerate triangles where the two vertices approach one another from different directions. There is still an exception in that the angle need not be a right angle, but we see that this can only occur when one of the edge lengths is zero - which is also, in some sense, exceptional.
Sep 7, 2021 at 17:53 comment added Robin Saunders I feel what this sort of exception is really telling us is that the situation we're interested in (triangles whose edges may or may not be perpendicular) calls for a different definition of the underlying object. I'm not saying "the correct definition of triangle has three distinct points", or anything along those lines. Rather, there are several potential definitions, which in the generic case are equivalent but have different behaviour for edge cases (no pun intended).
Sep 7, 2021 at 13:22 comment added Alexandre Eremenko If $A=B$ then $ABC$ is not a triangle.
Sep 7, 2021 at 8:48 history answered bathalf15320 CC BY-SA 4.0