Ceva's Theorem

Ceva's Theorem states that if three lines are drawn in a triangle from each vertex to the opposite sides (AA', BB', and CC' in the figure) they intersect in a single point if, and only if, the sides are divided into parts so that : .

  The theorem is named for Giovanni Ceva, an Italian mathematician who lived from 1648 to 1734. The lines from each vertex to the opposite side are often called Cevians in his honor.  You can find a biography of Ceva at the St. Andrews University web site.

This theorem makes some of the geometric proofs  of concurrency almost trivial corollarys.   The medians, for example, divide each side into a 1:1 ratio, so that all three of the ratios in the formula equal 1, and therefore have a product of one.  It is almost as easy to prove the angle bisectors meet in a single point with Ceva's theorem.

 

Here you can find a clever javascript proof of Ceva's Thm that requires nothing beyond middle school geometry formulas.

There is a second simple identity that is known, but not WELL known. Let three cevians be drawn from the vertices (A, B, and C)through a common point, P, and intersecting the opposite sides (perhaps extended) at A', B', and C' as in the figure. Then for the points as described, it is true that AP/AA' + BP/BB' + CP/CC' = 2 . I was first exposed to this pretty little property in a note to the MathForum Geometry discussion list by the Greek Mathematician Antreas P. Hatzipolakis.

I recently learned on one of the geometry discussion lists at the Math Forum that the Cevian is also used in 3-D for the segment from a vertex of a tetrahedron to the opposite face (possibly extended). In the same thread I had speculated that I thought the property above would extend to the tetrahedron as well with a sum of the ratios equal to three. Eisso J Atzema of The University of Maine confirmed my belief with a simple proof that extended from triangles to any N-dimensional simplex. I quote directly from his post:

The easiest proof probably is to note that the simplex A1...An-1P shares an n-1-dimensional base with A1...An, while its height equals PAn'/AnAn' times the 'height' of A1..An. Consequently, the volume of A1..AnO is PAn'/AnAn' times the volume of A1..An. A similar argument applies to any simplex formed by replacing a vertex of A1..An by P. Now the volume of all these simplices sums up to the volume of A1..An, from which the relation immediately follows.
. He added that he is co-authoring a paper on the history of this lemma and some applications. If I receive notice of the publication, a citation will be added.