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This theorem was proved by Giovanni Ceva (1648-1734).
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(The lines that meet at a point are said to be concurrent) |
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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, AGE and CBE, AGP and DBP, DCP and AHP. From these and in that order we derive the following proportions:
from the last two we conclude that AG/BD = AH/DC and, hence,
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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 P, 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 P is the point of intersection of BE and CF and draw the line AP 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. That is, D' and D are one and the same point.
Q.E.D.
The "area" form of Ceva's theorem is an immediate corollary, stating
that three cevians meet at a point iff the product of the ratios of the
areas Using the sine formula for the area of a triangle, equation 2 becomes The lengths of the sides of triangle ABC all cancel, leaving us with |
Internet references
http://www.cut-the-knot.com/Generalization/ceva.html has a java-enabled diagram on which you can drag the vertices of the triangle, and watch the various similar triangles in action.
Summary of geometrical theorems
Triangles -- a home page for anything about triangles.
Carnot's Theorem -- AC'2 - BC'2 + BA'2 - CA'2 + CB'2 - AB'2 = 0 iff perpendiculars from A', B', and C' concur on a point.
Isogonal Conjugates -- the relationship between the orthocenter and circumcenter of a triangle
Proof that the median and altitude of a right triangle are reflections about the bisector iff ABC is a right triangle.
Triangle Trisection -- If a point, P, on the median of triangle ABC is the isogonal conjugate of point Q, on the altitude of ABC, then ABC is a right triangle.
The webmaster and author of this Math Help site is Graeme McRae.
Ceva's theorem states that given three arbitrary cevians AD, BE and CF, the
three of them all meet at a point P if and only if
