Hello, and thanks for this wonderful site!There seems to be a confusion in the Barycenter section of the Ceva's theorem discussion (https://www.cut-the-knot.org/Generalization/ceva.shtml#barycenter). It is supposed to present an additional proof of Ceva's theorem, as it'says:
"Assuming that the barycenter's location is independent of the way it's computed we would actually get a different proof of the Ceva Theorem . . ."
However a little later Ceva's theorem is invoked:
"Ceva's theorem guarantees condition (1). From the foregoing discussion, there exist three masses . . ."
So it seems that it cannot constitute a proof of the theorem, if it relies on that very same theorem!
The needed correction seems to be straightforward. On the one hand, given Ceva's equation:
(1) AF/FB * BD/DC * CE/EA = 1,
one can define the masses wA, wB, wC as suggested (ibid), so that F becomes the barycenter of wA and wB, D of wB and wC, and E of wC and wA. Then it is guaranteed, by the properties of barycenters, that the common barycenter lies on CF, also on AD, also on BE. And since there is one and only one barycenter, it is guaranteed that these three Cevians are concurrent.
On the other hand, given three concurrent Cevians AD, BE and CF, one can define again three masses wA, wB, wC as suggested (ibid), in such a way that, say, F is set to be the barycenter of wA and wB, and D the barycenter of wB and wC. It is at this stage not guaranteed that E is the barycenter of wC and wA. However, the common barycenter (of wA, wB and wC) is guaranteed to lie, by the properties of barycenters, on CF as well as on AD, so it must lie on their intersection, which is also their intersection with BE (call it K). Now define E' on CA such that it is the barycenter of wA and wC. The common barycenter must lie on BE'. But we already concluded that the common barycenter is K. So K lies on BE'. Therefore BE' = BE (these two segments share two distinct points, B and K). Therefore E' = E. And this immediately entails Ceva's equation (1).
Ram
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