# Difference between revisions of "Ceva's Theorem"

(restated theorem in conventional (and, in my opinion, superior) way; rewrote first proof; also wrote proof for trig version (I think Law of Sines is a dead end)) |
(→Trigonometric Form: typos) |
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and similarly, | and similarly, | ||

<center> | <center> | ||

− | <math> \frac{[CBX]}{[XBA]} = \frac{ | + | <math> \frac{[CBX]}{[XBA]} = \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} ;\; \frac{[ACX]}{[XCB]} = \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math>. </center> |

It follows that | It follows that | ||

<center> | <center> | ||

− | <math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{ | + | <math> \frac{\sin BAD}{\sin DAC} \cdot \frac{\sin CBE}{\sin EBA} \cdot \frac{\sin ACF}{\sin FCB} = \frac{AB \cdot \sin BAD}{AC \cdot \sin DAC} \cdot \frac{BC \cdot \sin CBE}{BA \cdot \sin EBA} \cdot \frac{CA \cdot \sin ACF}{CB \cdot \sin FCB} </math> <br> <br> <math> \qquad = \frac{[BAX]}{[XAC]} \cdot \frac{[CBX]}{[XBA]} \cdot \frac{[ACX]}{[XCB]} = 1 </math>. |

</center> | </center> | ||

+ | |||

+ | Here, sign is irrelevant, as we may interpret the sines of [[directed angles]] mod <math> \displaystyle \pi </math> 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. {{Halmos}} | The converse follows by an argument almost identical to that used for the first form of Ceva's Theorem. {{Halmos}} |

## Revision as of 07:55, 8 June 2007

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

## Statement

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

,

where lengths are directed.

(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. ∎

## 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 . - 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.