1||0|76|0|
0|0|0|||||On the barycenter proof for Ceva%27s theorem|ram||11:29:33|07/06/2009|Hello%2C and thanks for this wonderful site%21%0D%0A%0D%0AThere seems to be a confusion in the Barycenter section of the Ceva%27s theorem discussion %28%5Ba%5Dhttp%3A%2F%2Fwww.cut-the-knot.org%2FGeneralization%2Fceva.shtml%23barycenter%5B%2Fa%5D%29. It is supposed to present an additional proof of Ceva%27s theorem%2C as it says%3A%0D%0A%0D%0A%22Assuming that the barycenter%27s location is independent of the way it%27s computed we would actually get a different proof of the Ceva Theorem . . .%22%0D%0A%0D%0AHowever a little later Ceva%27s theorem is invoked%3A%0D%0A%0D%0A%22Ceva%27s theorem guarantees condition %281%29. From the foregoing discussion%2C there exist three masses . . .%22%0D%0A%0D%0ASo it seems that it cannot constitute a proof of the theorem%2C if it relies on that very same theorem%21%0D%0A%0D%0AThe needed correction seems to be straightforward. On the one hand%2C given Ceva%27s equation%3A%0D%0A%281%29 AF%2FFB %2A BD%2FDC %2A CE%2FEA %3D 1%2C%0D%0Aone can define the masses w%5Bsub%5DA%5B%2Fsub%5D%2C w%5Bsub%5DB%5B%2Fsub%5D%2C w%5Bsub%5DC%5B%2Fsub%5D as suggested %28ibid%29%2C so that F becomes the barycenter of w%5Bsub%5DA%5B%2Fsub%5D and w%5Bsub%5DB%5B%2Fsub%5D%2C D of w%5Bsub%5DB%5B%2Fsub%5D and w%5Bsub%5DC%5B%2Fsub%5D%2C and E of w%5Bsub%5DC%5B%2Fsub%5D and w%5Bsub%5DA%5B%2Fsub%5D. Then it is guaranteed%2C by the properties of barycenters%2C that the common barycenter lies on CF%2C also on AD%2C also on BE. And since there is one and only one barycenter%2C it is guaranteed that these three Cevians are concurrent.%0D%0A%0D%0AOn the other hand%2C given three concurrent Cevians AD%2C BE and CF%2C one can define again three masses w%5Bsub%5DA%5B%2Fsub%5D%2C w%5Bsub%5DB%5B%2Fsub%5D%2C w%5Bsub%5DC%5B%2Fsub%5D as suggested %28ibid%29%2C in such a way that%2C say%2C F is set to be the barycenter of w%5Bsub%5DA%5B%2Fsub%5D and w%5Bsub%5DB%5B%2Fsub%5D%2C and D the barycenter of w%5Bsub%5DB%5B%2Fsub%5D and w%5Bsub%5DC%5B%2Fsub%5D. It is at this stage not guaranteed that E is the barycenter of w%5Bsub%5DC%5B%2Fsub%5D and w%5Bsub%5DA%5B%2Fsub%5D. However%2C the common barycenter %28of w%5Bsub%5DA%5B%2Fsub%5D%2C w%5Bsub%5DB%5B%2Fsub%5D and w%5Bsub%5DC%5B%2Fsub%5D%29 is guaranteed to lie%2C by the properties of barycenters%2C on CF as well as on AD%2C so it must lie on their intersection%2C which is also their intersection with BE %28call it K%29. Now define E%27 on CA such that it is the barycenter of w%5Bsub%5DA%5B%2Fsub%5D and w%5Bsub%5DC%5B%2Fsub%5D. The common barycenter must lie on BE%27. But we already concluded that the common barycenter is K. So K lies on BE%27. Therefore BE%27 %3D BE %28these two segments share two distinct points%2C B and K%29. Therefore E%27 %3D E. And this immediately entails Ceva%27s equation %281%29.%0D%0A%0D%0ARam%0D%0A
1|1|0|||||RE%3A On the barycenter proof for Ceva%27s theorem|alexb||13:58:34|07/06/2009|Yes%2C thank you. %0D%0A%0D%0AThe page has been written piecemeal%2C in several stages%2C with several rearrangements and much copy-and-pasting.%0D%0A%0D%0AI have it upgraded. %0D%0A%0D%0AThank you again%2C%0D%0AAlex