Difference between revisions of "Ceva's Theorem"

Revision as of 16:40, 23 June 2006

Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.

Statement

http://billydorminy.homelinux.com/aopswiki/cevathm.png

A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that

$BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$

where all segments in the formula are directed segments.

Proof

Let ${X,Y,Z}$ be points on ${BC}, {CA}, {AB}$ respectively such that $AX,BY,CZ$ are concurrent, and let ${P}$ be the point where $AX$ , $BY$ and $CZ$ meet. Draw a parallel to $AB$ through the point ${C}$ . Extend $AX$ until it intersects the parallel at a point $\displaystyle{A'}$ . Construct $\displaystyle{B'}$ in a similar way extending $BY$ .

(ceva1.png)

The triangles $\displaystyle{\triangle{ABX}}$ and $\displaystyle{\triangle{A'CX}}$ are similar, and so are $\displaystyle\triangle{ABY}$ and $\triangle{CB'Y}$ . Then the following equalities hold:

$\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}$

and thus

$\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)$

Notice that if directed segments are being used, then $AB$ and $BA$ have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed $CA'$ to $A'C$ .

Now we turn to consider the following similarities: $\triangle{AZP}\sim\triangle{A'CP}$ and $\triangle BZP\sim\triangle B'CP$ . From them we get the equalities

$\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}$

which lead to

$\frac{AZ}{ZB}=\frac{A'C}{CB'}$ .

Multiplying the last expression with (1) gives

$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$

and we conclude the proof.

To prove the converse, suppose that ${X,Y,Z}$ are points on ${BC}, {CA}, {AB}$ respectively and satisfying

$\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.$

Let $Q$ be the intersection point of $AX$ with $BY$ , and let $Z'$ be the intersection of $CQ$ with $AB$ . Since then $AX,BY,CZ'$ are concurrent, we have

$\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$

and thus

$\frac{AZ'}{Z'B}=\frac{AZ}{ZB}$

which implies $Z=Z'$ , and therefore $AX,BY,CZ$ are concurrent.

(proof courtesy planetmath.org, used under GNU License)

Example

Suppose AB, AC, and BC have lengths 13, 14, and 15. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$ . Find BD and DC.

If $BD = x$ and $DC = y$ , then $10x = 40y$ , and ${x + y = 15}$ . From this, we find $x = 12$ and $y = 3$ .

@@ Line 6: / Line 6: @@
 A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that
-<br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br>
+<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot FB</math></center><br>
 where all segments in the formula are directed segments.

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Difference between revisions of "Ceva's Theorem"

Revision as of 16:40, 23 June 2006

Contents

Statement

Proof

Example

See also