# Difference between revisions of "Ceva's Theorem"

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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>\displaystyle{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>. | 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>\displaystyle{A'}</math>. Construct <math>\displaystyle{B'}</math> in a similar way extending <math>BY</math>. | ||

<center>''(ceva1.png)''</center> | <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: | + | The triangles <math>\displaystyle{\triangle{ABX}}</math> and <math>\displaystyle{\triangle{A'CX}}</math> are similar, and so are <math>\displaystyle\triangle{ABY}</math> and <math>\triangle{CB'Y}</math>. Then the following equalities hold: |

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

− | + | <br> | |

and thus | and thus | ||

− | <math> | + | <center><math>\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C} \qquad(1)</math></center> |

− | + | <br> | |

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>. | 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>. | ||

− | + | <br><br> | |

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

− | + | <br> | |

which lead to | which lead to | ||

− | <math> | + | <center><math>\frac{AZ}{ZB}=\frac{A'C}{CB'}</math>.</center> |

− | + | <br> | |

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

− | <math> | + | <center><math>\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1</math></center> |

− | + | <br> | |

and we conclude the proof. | and we conclude the proof. | ||

− | + | <br><br> | |

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

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

− | + | <br> | |

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

− | + | <br> | |

and thus | and thus | ||

− | <math> | + | <center><math>\frac{AZ'}{Z'B}=\frac{AZ}{ZB}</math></center> |

− | + | <br> | |

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

## Revision as of 17:01, 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 and are similar, and so are and . Then the following equalities hold:

and thus

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

which lead to

Multiplying the last expression with (1) gives

and we conclude the proof.

To prove the converse, suppose that are points on respectively and satisfying

Let be the intersection point of with , and let be the intersection of with . Since then are concurrent, we have

and thus

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 .