 Mathematics Magazine

Ceva's Theorem In a triangle ABC, three lines AD, BE and CF intersect at a single point K if and only if

 (1) AF/FB · BD/DC · CE/EA = 1

(The lines that meet at a point are said to be concurrent.)

Extend the lines BE and CF beyond the triangle until they meet GH, the line through A parallel to BC. There are several pairs of similar triangles: AHF and BCF, AEG and BCE, AGK and BDK, CDK and AHK. From these and in that order we derive the following proportions:

 AF/FB=AH/BC (*) CE/EA=BC/AG (*) AG/BD=AK/DK AH/DC=AK/DK

from the last two we conclude that AG/BD = AH/DC and, hence,

 BD/DC = AG/AH (*).

Multiplying the identities marked with (*) we get

 AF/FB · BD/DC · CE/EA = AH/BC · BC/AG · AG/AH = (AH·BC·AG)/(BC·AG·AH) = 1

Therefore, if the lines AD, BE and CF intersect at a single point K, the identity (1) does hold. Which is to say that the fact of the three lines intersecting at one point is sufficient for the condition (1) to hold. Let us now prove that it's also necessary. This would constitute the second part of the theorem. In other words, let us prove that if (1) holds then AD, BE, CF are concurrent.

Indeed, assume that K is the point of intersection of BE and CF and draw the line AK until its intersection with BC at a point D'. Then, from the just proven part of the theorem it follows that

 AF/FB · BD'/D'C · CE/EA = 1

On the other hand, it's given that

 AF/FB · BD/DC · CE/EA = 1

Combining the two we get

 BD'/D'C = BD/DC or BD'/D'C + 1= BD/DC + 1 or (BD'+D'C)/D'C = (BD+DC)/DC

Finally

 BC/D'C = BC/DC
which immediately implies D'C=DC
"Chance favors the prepared mind." - Louis Pasteur