# proof of Ceva’s theorem

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As in the article on Ceva’s Theorem, we will consider directed line segments.

Let $X$, $Y$ and $Z$ be points on $BC$, $CA$ and $AB$, respectively, such that $AX$, $BY$ and $CZ$ 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 ${A}^{\prime}$. Construct ${B}^{\prime}$ in a similar^{} way extending $BY$.

The triangles $\mathrm{\u25b3}ABX$ and $\mathrm{\u25b3}{A}^{\prime}CX$ are similar, and so are $\mathrm{\u25b3}ABY$ and $\mathrm{\u25b3}C{B}^{\prime}Y$. Then the following equalities hold:

$$\frac{BX}{XC}=\frac{AB}{C{A}^{\prime}},\frac{CY}{YA}=\frac{C{B}^{\prime}}{BA}$$ |

and thus

$$\frac{BX}{XC}\cdot \frac{CY}{YA}=\frac{AB}{C{A}^{\prime}}\cdot \frac{C{B}^{\prime}}{BA}=\frac{C{B}^{\prime}}{{A}^{\prime}C}.$$ | (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 $C{A}^{\prime}$ to ${A}^{\prime}C$.

Now we turn to consider the following similarities: $\mathrm{\u25b3}AZP\sim \mathrm{\u25b3}{A}^{\prime}CP$ and $\mathrm{\u25b3}BZP\sim \mathrm{\u25b3}{B}^{\prime}CP$. From them we get the equalities

$$\frac{CP}{ZP}=\frac{{A}^{\prime}C}{AZ},\frac{CP}{ZP}=\frac{C{B}^{\prime}}{ZB}$$ |

which lead to

$$\frac{AZ}{ZB}=\frac{{A}^{\prime}C}{C{B}^{\prime}}.$$ |

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 ${Z}^{\prime}$ be the intersection of $CQ$ with $AB$. Since then $AX,BY,C{Z}^{\prime}$ are concurrent, we have

$$\frac{A{Z}^{\prime}}{{Z}^{\prime}B}\cdot \frac{BX}{XC}\cdot \frac{CY}{YA}=1$$ |

and thus

$$\frac{A{Z}^{\prime}}{{Z}^{\prime}B}=\frac{AZ}{ZB}$$ |

which implies $Z={Z}^{\prime}$, and therefore $AX,BY,CZ$ are concurrent.

Title | proof of Ceva’s theorem |
---|---|

Canonical name | ProofOfCevasTheorem |

Date of creation | 2013-03-22 12:38:54 |

Last modified on | 2013-03-22 12:38:54 |

Owner | yark (2760) |

Last modified by | yark (2760) |

Numerical id | 11 |

Author | yark (2760) |

Entry type | Proof |

Classification | msc 51A05 |