# Difference between revisions of "Ceva's Theorem"

Line 24: | Line 24: | ||

<math>\mathcal{QED}</math> | <math>\mathcal{QED}</math> | ||

+ | == Alternate Formulation == | ||

+ | |||

+ | The [[trig]] version of Ceva's Theorem states that cevians <math>AD,BE,CF</math> are concurrent if and only if | ||

+ | |||

+ | <center>$\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA.$</center> | ||

+ | |||

+ | === Proof === | ||

+ | |||

+ | ''This proof is incomplete. If you can finish it, please do so. Thanks!'' | ||

+ | |||

+ | We will use Ceva's Theorem in the form that was already proven to be true. | ||

+ | |||

+ | First, we show that if $\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA$, holds true then $BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$ which gives that the cevians are concurrent by Ceva's Theorem. The [[Law of Sines]] tells us that <center>$\frac{BD}{\sin BAD} = \frac{AB}{\sin ADB} \Leftrightarrow \sin BAD = \frac{BD}{AB\sin ADB}.$</center> | ||

+ | |||

+ | Likewise, we get | ||

+ | |||

+ | {| class="wikitable" style="margin: 1em auto 1em auto" | ||

+ | |- | ||

+ | | $\sin ACF = \frac{AF}{AC\sin CFA}$ | ||

+ | |- | ||

+ | | $\sin CBE = \frac{CE}{BC\sin BEC}$ | ||

+ | |- | ||

+ | | $\sin CAD = \frac{CD}{AC\sin ADC}$ | ||

+ | |- | ||

+ | | $\sin BCF = \frac{BF}{BC\sin BFC}$ | ||

+ | |- | ||

+ | | $\sin ABE = \frac{AE}{AB\sin AEB}$ | ||

+ | |} | ||

+ | |||

+ | Thus | ||

+ | |||

+ | {| class="wikitable" | ||

+ | |- | ||

+ | | <math>\sin BAD \sin ACF \sin CBE = \sin CAD \sin BCF \sin ABE</math> | ||

+ | |- | ||

+ | | $\frac{BD}{AB\sin ADB} \cdot \frac{AF}{AC\sin CFA} \cdot \frac{CE}{BC\sin BEC} = \frac{CD}{AC\sin ADC} \cdot \frac{BF}{BC\sin BFC} \cdot \frac{AE}{AB\sin AEB} | ||

+ | |} | ||

== Examples == | == Examples == |

## Revision as of 10:26, 22 December 2006

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

## Statement

A necessary and sufficient condition for where and are points of the respective side lines of a triangle , to be concurrent is that

where all segments in the formula are directed segments.

## Proof

Letting the altitude from to have length we have and where the brackets represent area. Thus . In the same manner, we find that . Thus

Likewise, we find that

Thus

## Alternate Formulation

The trig version of Ceva's Theorem states that cevians are concurrent if and only if

### Proof

*This proof is incomplete. If you can finish it, please do so. Thanks!*

We will use Ceva's Theorem in the form that was already proven to be true.

First, we show that if $\sin BAD \sin ACF \sin CBE = \sin DAC \sin FCB \sin EBA$, holds true then $BD\cdot CE\cdot AF = DC \cdot EA \cdot FB$ which gives that the cevians are concurrent by Ceva's Theorem. The Law of Sines tells us that

Likewise, we get

$\sin ACF = \frac{AF}{AC\sin CFA}$ |

$\sin CBE = \frac{CE}{BC\sin BEC}$ |

$\sin CAD = \frac{CD}{AC\sin ADC}$ |

$\sin BCF = \frac{BF}{BC\sin BFC}$ |

$\sin ABE = \frac{AE}{AB\sin AEB}$ |

Thus

$\frac{BD}{AB\sin ADB} \cdot \frac{AF}{AC\sin CFA} \cdot \frac{CE}{BC\sin BEC} = \frac{CD}{AC\sin ADC} \cdot \frac{BF}{BC\sin BFC} \cdot \frac{AE}{AB\sin AEB} |

## Examples

- 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 . - See the proof of the concurrency of the altitudes of a triangle at the orthocenter.
- See the proof of the concurrency of the perpendicual bisectors of a triangle at the circumcenter.