Ceva's theorem

From Academic Kids

Ceva's Theorem (pronounced "Cheva") is a very popular theorem in elementary geometry. Given a triangle ABC, and points D, E, and F that lie on lines BC, CA, and AB respectively, the theorem states that lines AD, BE and CF are concurrent if and only if

It was first proved by Giovanni Ceva.

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Cevastheorem.jpg

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Proof

Suppose <math>AD<math>, <math>BE<math> and <math>CF<math> intersect at a point <math>X<math>. Because <math>\triangle BXD<math> and <math>\triangle CXD<math> have the same height, we have

<math>\frac{|\triangle BXD|}{|\triangle CXD|}=\frac{BD}{DC}.<math>

Similarly,

<math>\frac{|\triangle BAD|}{|\triangle CAD|}=\frac{BD}{DC}.<math>

From this it follows that

<math>\frac{BD}{DC}=

\frac{|\triangle BAD|-|\triangle BXD|}{|\triangle CAD|-|\triangle CXD|}

=\frac{|\triangle ABX|}{|\triangle CAX|}.<math>

Similarly,

<math>\frac{CE}{EA}=\frac{|\triangle BCX|}{|\triangle ABX|}<math>, and
<math>\frac{AF}{FB}=\frac{|\triangle CAX|}{|\triangle BCX|}<math>.

Multiplying these three equations gives

as required. Conversely, suppose that the points <math>D<math>, <math>E<math> and <math>F<math> satisfy the above equality. Let <math>AD<math> and <math>BE<math> intersect at <math>X<math>, and let <math>CX<math> intersect <math>AB<math> at <math>F'<math>. By the direction we have just proven,

Comparing with the above equality, we obtain

Adding 1 to both sides and using <math>AF'+F'B=AF+FB=AB<math>, we obtain

Thus <math>F'B=FB<math>, so that <math>F<math> and <math>F'<math> coincide (recalling that the distances are directed). Therefore <math>AD<math>, <math>BE<math> and <math>CF<math>=<math>CF'<math> intersect at <math>X<math>, and both implications are proven.

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External links

Ceva's Theorem, Interactive proof with animation and key concepts (http://agutie.homestead.com/files/ceva.htm) by Antonio Gutierrez from the land of the Incas
Derivations and applications of Ceva's Theorem (http://www.cut-the-knot.org/Generalization/ceva.shtml)
Cevian Nest (http://www.cut-the-knot.org/Curriculum/Geometry/CevaNest.shtml)
Trigonometric Form of Ceva's Theorem (http://www.cut-the-knot.org/triangle/TrigCeva.shtml)de:Satz von Ceva