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Owner confidence rating: High Entry average rating: No information on entry rating
Ceva's theorem (Theorem)

Let $ABC$ be a given triangle and $P$ any point of the plane. If $X$ is the intersection point of $AP$ with $BC$, $Y$ the intersection point of $BP$ with $CA$ and $Z$ is the intersection point of $CP$ with $AB$, then

\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}

Conversely, if $X,Y,Z$ are points on $BC,CA,AB$ respectively, and if

\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}

then $AX,BY,CZ$ are concurrent.

Remarks: All the segments are directed segments (that is $AB=-BA$), and so theorem is valid even if the points $X,Y,Z$ are in the prolongations (even at the infinity) and $P$ is any point on the plane (or at the infinity).

\includegraphics[scale=0.75]{ceva}



"Ceva's theorem" is owned by drini. [ owner history (1) ]
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See Also: triangle, median, centroid, orthocenter, orthic triangle, cevian, incenter, Gergonne point, Menelaus' theorem, proof of Van Aubel theorem, Van Aubel theorem, bisectors theorem, directed segment

Keywords:  Concurrence, Menelaus Theorem

Attachments:
proof of Ceva's theorem (Proof) by yark
trigonometric version of Ceva's theorem (Theorem) by drini
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Cross-references: infinity, even, valid, theorem, directed segments, segments, concurrent, conversely, intersection, plane, point, triangle
There are 4 references to this entry.

This is version 12 of Ceva's theorem, born on 2001-11-09, modified 2005-01-30.
Object id is 745, canonical name is CevasTheorem.
Accessed 8447 times total.

Classification:
AMS MSC: 51A05 (Geometry :: Linear incidence geometry :: General theory and projective geometries)
Pending Errata and Addenda
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Discussion
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Generalizations of Ceva's Theorem by zorba on 2006-11-18 11:25:30
I discovered an article with Generalizations of Ceva's Theorem
in arXiv at:
http://xxx.lanl.gov/pdf/math.GM/0610689
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