# Difference between revisions of "Ceva's Theorem"

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

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

=== Proof === | === Proof === | ||

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

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

Likewise, we get | Likewise, we get | ||

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

|- | |- | ||

− | | | + | | <math>\sin ACF = \frac{AF}{AC\sin CFA}</math> |

|- | |- | ||

− | | | + | | <math>\sin CBE = \frac{CE}{BC\sin BEC}</math> |

|- | |- | ||

− | | | + | | <math>\sin CAD = \frac{CD}{AC\sin ADC}</math> |

|- | |- | ||

− | | | + | | <math>\sin BCF = \frac{BF}{BC\sin BFC}</math> |

|- | |- | ||

− | | | + | | <math>\sin ABE = \frac{AE}{AB\sin AEB}</math> |

|} | |} | ||

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

|- | |- | ||

− | | | + | | <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}</math> |

|} | |} | ||

## Revision as of 15:43, 24 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 , holds true then which gives that the cevians are concurrent by Ceva's Theorem. The Law of Sines tells us that

Likewise, we get

Thus

## 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.