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Ceva's theorem

(Redirected from Ceva's Theorem)

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

\frac{AF}{FB}  \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.

It was first proved by Giovanni Ceva.


Proof

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

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

Similarly,

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

From this it follows that

\frac{BD}{DC}= \frac{|\triangle BAD|-|\triangle BXD|}{|\triangle CAD|-|\triangle CXD|} =\frac{|\triangle ABX|}{|\triangle CAX|}.

Similarly,

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

Multiplying these three equations gives

\frac{AF}{FB}  \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1

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

\frac{AF'}{F'B}  \cdot \frac{BD}{DC} \cdot \frac{CE}{EA} = 1.

Comparing with the above equality, we obtain

\frac{AF'}{F'B}=\frac{AF}{FB}.

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

\frac{AB}{F'B}=\frac{AB}{FB}.

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

See also

External links



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01-04-2007 01:21:04