# Difference between revisions of "Ceva's Theorem"

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== Problems == | == Problems == | ||

===Introductory=== | ===Introductory=== | ||

− | *Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>, find <math>BD</math> and <math>DC</math>. ([[ | + | *Suppose <math>AB, AC</math>, and <math>BC</math> have lengths <math>13, 14</math>, and <math>15</math>, respectively. If <math>\frac{AF}{FB} = \frac{2}{5}</math> and <math>\frac{CE}{EA} = \frac{5}{8}</math>, find <math>BD</math> and <math>DC</math>. ([[Ceva's Theorem/Problems|Source]]) |

+ | |||

===Intermediate=== | ===Intermediate=== | ||

===Olympiad=== | ===Olympiad=== |

## Revision as of 22:42, 14 November 2007

**Ceva's Theorem** is a criterion for the concurrence of cevians in a triangle.

## Contents

## Statement

Let be a triangle, and let be points on lines , respectively. Lines concur iff (if and only if)

,

where lengths are directed. This also works for the reciprocal or each of the ratios, as the reciprocal of is .

(Note that the cevians do not necessarily lie within the triangle, although they do in this diagram.)

## Proof

We will use the notation to denote the area of a triangle with vertices .

First, suppose meet at a point . We note that triangles have the same altitude to line , but bases and . It follows that . The same is true for triangles , so

Similarly, and , so

.

Now, suppose satisfy Ceva's criterion, and suppose intersect at . Suppose the line intersects line at . We have proven that must satisfy Ceva's criterion. This means that

so

and line concurrs with and . ∎

## Trigonometric Form

The trigonometric form of Ceva's Theorem (Trig Ceva) states that cevians concur if and only if

### Proof

First, suppose concur at a point . We note that

and similarly,

It follows that

.

Here, sign is irrelevant, as we may interpret the sines of directed angles mod to be either positive or negative.

The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. ∎

## Problems

### Introductory

- Suppose , and have lengths , and , respectively. If and , find and . (Source)

### Intermediate

### Olympiad

## Other Notes

- The concurrence of the altitudes of a triangle at the orthocenter and the concurrence of the perpendicual bisectors of a triangle at the circumcenter can both be proven by Ceva's Theorem (the latter is a little harder). Furthermore, the existance of the centroid can be shown by Ceva, and the existance of the incenter can be shown using trig Ceva. However, there are more elegant methods for proving each of these results, and in any case, any result obtained by classic Ceva's Theorem can be proven using ratios of areas.
- The existance of isotonic conjugates can be shown by classic Ceva, and the existance of isogonal conjugates can be shown by trig Ceva.