Difference between revisions of "Ceva's Theorem"

m (\cdot!!)
(Statement)
Line 5: Line 5:
 
http://billydorminy.homelinux.com/aopswiki/cevathm.png
 
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
+
A [[necessary and sufficient]] condition for <math>AD, BE, CF,</math> where <math>D, E,</math> and <math>F</math> are points of the respective side lines <math>BC, CA, AB</math> of a triangle <math>ABC</math>, to be concurrent is that
 
<br><center><math>BD\cdot CE\cdot AF = DC \cdot EA \cdot 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.
+
where all segments in the formula are [[directed segments]].
  
 
== Proof ==
 
== Proof ==

Revision as of 10:21, 18 August 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 $y = 3$.

See also

Invalid username
Login to AoPS