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The intersection at one point (called Gergonne point) of the lines from vertices of a triangle to contact points of the inscribed circle can be proved immediately using Ceva's theorem.

Is there a direct proof that does not pass through Ceva's formula?

Edit: I am hoping for a metric Euclidean proof using lengths and angles but this should not limit the answers. If you can prove it using algebraic K theory, go ahead.

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  • $\begingroup$ In case it saves time for search engine aficionados, I did try that in several variations. There is a large number of web pages with the usual Ceva proof that the question seeks to avoid. $\endgroup$ – zyx Mar 4 '13 at 21:24
  • $\begingroup$ What's wrong with Ceva? (Just being curious.) $\endgroup$ – dtldarek Mar 4 '13 at 21:29
  • $\begingroup$ Nothing. This is a methodological question. $\endgroup$ – zyx Mar 4 '13 at 21:49
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It is a conclusion of a degenerated case of the Brianchon's theorem:

For a hexagon $ABCDEF$ with an inscribed conic section the lines $AD$, $BE$, $CF$ coincide in a common point.

Just set $A$, $C$ and $E$ to be vertices of the triangle, and $B$, $D$ and $F$ the points where the incircle touches the triangle. Some immediate conclusion: the Gergonne point could be generalized to ellipse and other conics. Quick search reveals this idea is not new and there is even some literature on it, e.g. this nice pdf and also this page.

I hope this is what you were looking for ;-)

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  • $\begingroup$ Thank you (+1). It is a bit late to clarify, but I am looking for a metric equality-chasing proof in classical Euclidean style. $\endgroup$ – zyx Mar 4 '13 at 22:52
  • $\begingroup$ Also, Ceva theorem is itself a degeneration of projective theorems on hexagons, so in a sense this is just a deformation of the usual proof. $\endgroup$ – zyx Mar 4 '13 at 22:54
  • $\begingroup$ Well, Ceva does involve the formula-checking, and I thought you wanted to avoid this particular thing. Anyway, if I were you, I would do a quick search through Pascal's theorem proofs, just in case. Again, you could say that Pascal is multiple Menelaus so this is similar of even worse than inlining the Ceva's proof, but still, maybe you will find something interesting? ;-) $\endgroup$ – dtldarek Mar 4 '13 at 23:01
  • $\begingroup$ For example, if there is some characterization of each line as a locus of points $p$ where $f(p,A_i)=f(p,A_j)$ and the triangle is $A_1 A_2 A_3$, this would prove concurrency. I don't mind checking formulas but I wonder if some insight is available as in the other non-Ceva proofs. $\endgroup$ – zyx Mar 4 '13 at 23:06
  • $\begingroup$ mathworld.wolfram.com/BrianchonPoint.html . List of "Brianchon points" of "inconics"! $\endgroup$ – zyx Mar 19 '13 at 5:57

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