Talk:Ceva's theorem

Popular how?

In what way could this theorem be construed to be "popular" among schoolchildren? :)

I popped in to ask much the same question...perhaps they're defining "schoolchildren" to be "high school and college students who are fanatically interested in geometry" Kaz 01:15, 5 Feb 2005 (UTC)

I think it's a) popular to teach it b) has been frequently examined historically. Mr. Jones 10:50, 5 May 2005 (UTC)[reply]

Your explanation would be benefitted enormously if you would include the original triangle when you begin your first proof via area and you color code the various areas.

Show all the steps used to arrive at your conclusion. This is not a book, computer screen space does not cost money.

Always, Always, Always be as clear as possible as you don't know who will be reading your proofs. — Preceding unsigned comment added by 76.178.133.207 (talk) 07:05, 25 March 2020 (UTC)[reply]

Errors ?

There are many errors in this article in case 2 (D or E or F not on the line segments)

The egality is also true if (AD), (BE), (CF) are parallel
To proove that the line are concurrent or parallel, the egality must be writen with measurements of line segments that are allowed negative values. So the proof is'nt valable.
${\frac {AF'}{F'B}}={\frac {AF}{FB}}$ ==> F=F' or F and F' are harmonic division of [AB]

I can't modify this article because I'm french and I can't write in english. HB on fr 7 August 2007 (UTC)

2 cases

As I see it, the proof doesn't split into 2 cases, yet the pictures suggest so. —Preceding unsigned comment added by Richard Molnár-Szipai (talk • contribs) 14:39, 11 March 2008 (UTC)[reply]

Generalisations?

There's one type of generalisation we're familiar with, that is, polygons with odd number of sides. Are there any other types of generalisations? —Preceding unsigned comment added by 117.195.34.180 (talk) 14:33, 16 July 2008 (UTC)[reply]

duality?

In what sense is this theorem the "dual" of Menelaus's theorem? Both Ceva and Menelaus employ distances which are not projective invariants. Neither theorem is a theorem in projective geometry. Tkuvho (talk) 11:07, 1 June 2010 (UTC)[reply]

The figures for the two theorem are dual in the projective sense [Meneleus: a triangle and the intersections of a line with its edges ; Ceva: a triangle and the lines joining a point to its vertices]. Both theorems have projective generalizations that are dual - they are statements about "cross-ratios" Ericlord (talk) 20:45, 15 February 2012 (UTC)[reply]

The article doesn't say they are duals any more. One is a statement about complete quadrilaterals and the other is about quadrangles and these are indeed dual figures. The projective generalization should probably be added to the article at some point; I didn't see many sources when I looked though. There are also spherical and hyperbolic versions of both theorems but again sources are rather scarce.--RDBury (talk) 01:56, 16 February 2012 (UTC)[reply]

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Tetrahedron

What form has the statement of the theorem in the case of a tetrahedron?--82.137.11.174 (talk) 22:23, 8 April 2017 (UTC)[reply]

Presented proof academically explicitly alleged as flawed since November 2014

While researching something almost entirely unrelated, I serendipitously came across the following paper (hence, I only briefly checked it): Bantchev, Boyko B. (April 2–6, 2015). How not to fail proving Ceva’s theorem (PDF). Proceedings of the Forty Fourth Spring Conference of the Union of Bulgarian Mathematicians. Matematika i matematičesko obrazovanie. Vol. 2015. SOK “Kamchia”. pp. 249–257. ISSN 1313-3330.

The paper alleges explicitly the proof in this article(as of some revision viewable in November 2014) as flawed. Reviewing the article history, I don't see any notable changes in the proof since, as such, I'd assume the proof as incorrect, but I lack both the time (it already took QUITE some time to dig out what seems like the correct metadata for the above citation) & expertise to check it (& the paper) in detail. As such, I add this talk section & the disputed & expert attention templates to the article. --No identd (talk) 13:19, 7 January 2019 (UTC)[reply]

This paper is not really a reliable source. Moreover, it asserts that many proofs are flawed, without saying anything about the nature of the flaw. On the other hand, the proof that is given here is properly sourced, and an independent source for several other proofs are given.

I have read the proof given in the article. It seems to be not flawed. The only issue is that it involves several cases, and these cases are not always listed explicitly (there are 7 cases for the position of the point O, that can be reduced to three up to a circular permutation of the vertices). Thus a proof that is not case dependent is somehow better, and this is the object of the cited article. But this does not make wrong a case-dependent proof.

Therefore, I'll remove the tags, leaving this thread for a further discussion, if needed. D.Lazard (talk) 14:53, 7 January 2019 (UTC)[reply]