# Difference between revisions of "Ceva's Theorem"

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<br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br> | <br><center><math>BD * CE * AF = +DC * EA * FB</math></center><br> | ||

where all segments in the formula are directed segments. | where all segments in the formula are directed segments. | ||

+ | |||

+ | == Proof == | ||

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

+ | <center>''(ceva1.png)''</center> | ||

+ | The triangles <math>\triangle{ABX}<math> and <math>\triangle{A'CX}</math> are similar, and so are <math>\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold: | ||

+ | <math>\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}</math> | ||

+ | |||

+ | and thus | ||

+ | <math>\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)</math> | ||

+ | |||

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

+ | |||

+ | Now we turn to consider the following similarities: <math>\triangle{AZP}\sim\triangle{A'CP}</math> and <math>\triangle BZP\sim\triangle B'CP</math>. From them we get the equalities | ||

+ | <math>\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}</math> | ||

+ | |||

+ | which lead to | ||

+ | <math>\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}</math> | ||

+ | |||

+ | Multiplying the last expression with (1) gives | ||

+ | <math>\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}</math> | ||

+ | |||

+ | and we conclude the proof. | ||

+ | |||

+ | To prove the converse, suppose that <math>X,Y,Z</math> are points on <math>{BC, CA, AB}</math> respectively and satisfying | ||

+ | <math>\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}</math> | ||

+ | |||

+ | Let <math>Q</math> be the intersection point of <math>AX</math> with <math>BY</math>, and let <math>Z'</math> be the intersection of <math>CQ</math> with <math>AB</math>. Since then <math>AX,BY,CZ'</math> are concurrent, we have | ||

+ | <math>\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}</math> | ||

+ | |||

+ | and thus | ||

+ | <math>\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}</math> | ||

+ | |||

+ | which implies <math>Z=Z'</math>, and therefore <math>AX,BY,CZ</math> are concurrent. | ||

== Example == | == Example == |

## Revision as of 15:36, 20 June 2006

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

## Contents

## Statement

*(awaiting image)*

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

where all segments in the formula are directed segments.

## Proof

Let be points on respectively such that are concurrent, and let be the point where , and meet. Draw a parallel to through the point . Extend until it intersects the parallel at a point . Construct in a similar way extending .

*(ceva1.png)*

The triangles are similar, and so are and . Then the following equalities hold: $\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and thus $\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

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

Now we turn to consider the following similarities: and . From them we get the equalities $\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

which lead to $\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

Multiplying the last expression with (1) gives $\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and we conclude the proof.

To prove the converse, suppose that are points on respectively and satisfying $\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

Let be the intersection point of with , and let be the intersection of with . Since then are concurrent, we have $\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and thus $\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

which implies , and therefore are concurrent.

## Example

Suppose AB, AC, and BC have lengths 13, 14, and 15. If and . Find BD and DC.

If and , then , and . From this, we find and .